e-Notes by Prof. P.

Nagaraju, RVCE, Bangalore

Course: Digital Communication (EC61)

Course instructors:

1. Mr. H. V.KumaraSwamy, RVCE, Bangalore

2. Mr. P.Nagaraju, RVCE, Bangalore
3. Ms. M.N.Suma, BMSCE, Bangalore

TEXT BOOK:
“Digital Communications”
Author: Simon Haykin Pub: John Wiley Student Edition, 2003

Reference Books:

1. “Digital and Analog Communication Systems” –

K. Sam Shanmugam, John Wiley, 1996.
2. “An introduction to Analog and Digital Communication”-
Simon Haykin, John Wiley, 2003
3. “Digital Communication- Fundamentals & Applications” –
Bernard Sklar, Pearson Education, 2002.
4. “Analog & Digital Communications”-
HSU, Tata Mcgraw Hill, II edition
 Chapter-1: Introduction

The purpose of a Communication System is to transport an information bearing signal

from a source to a user destination via a communication channel.

MODEL OF A COMMUNICATION SYSTEM

Information I/P Signal

Source and TRANSMITTER
Input
Transducer CHANNEL

O/P Signal
Destination
and Output RECEIVER
Transducer

Fig. 1.1: Block diagram of Communication System.

The three basic elements of every communication systems are Transmitter,

Receiver and Channel.
The Overall purpose of this system is to transfer information from one point
(called Source) to another point, the user destination.
The message produced by a source, normally, is not electrical. Hence an input
transducer is used for converting the message to a time – varying electrical quantity
called message signal. Similarly, at the destination point, another transducer converts the
electrical waveform to the appropriate message.
The transmitter is located at one point in space, the receiver is located at some
other point separate from the transmitter, and the channel is the medium that provides the
electrical connection between them.
The purpose of the transmitter is to transform the message signal produced by the
source of information into a form suitable for transmission over the channel.
The received signal is normally corrupted version of the transmitted signal, which
is due to channel imperfections, noise and interference from other sources.
The receiver has the task of operating on the received signal so as to reconstruct a
recognizable form of the original message signal and to deliver it to the user destination.

Communication Systems are divided into 3 categories:

1. Analog Communication Systems are designed to transmit analog information
using analog modulation methods.
2. Digital Communication Systems are designed for transmitting digital information
using digital modulation schemes, and
3. Hybrid Systems that use digital modulation schemes for transmitting sampled and
quantized values of an analog message signal.
ELEMENTS OF DIGITAL COMMUNICATION SYSTEMS:

The figure 1.2 shows the functional elements of a digital communication system.
Source of Information: 1. Analog Information Sources.
2. Digital Information Sources.

Analog Information Sources → Microphone actuated by a speech, TV Camera scanning a

scene, continuous amplitude signals.

Digital Information Sources → These are teletype or the numerical output of computer
which consists of a sequence of discrete symbols or letters.

An Analog information is transformed into a discrete information through the

process of sampling and quantizing.

Digital Communication System

Source of Source Channel

Information Encoder Encoder Modulator
Wave fo

Binary Stream Channel

Received Signal

User of Source Channel

Demodulator
Information Decoder Decoder

Fig 1.2: Block Diagram of a Digital Communication System

SOURCE ENCODER / DECODER:

The Source encoder ( or Source coder) converts the input i.e. symbol sequence
into a binary sequence of 0’s and 1’s by assigning code words to the symbols in the input
sequence. For eg. :-If a source set is having hundred symbols, then the number of bits
used to represent each symbol will be 7 because 27=128 unique combinations are
available. The important parameters of a source encoder are block size, code word
lengths, average data rate and the efficiency of the coder (i.e. actual output data rate
compared to the minimum achievable rate)
 At the receiver, the source decoder converts the binary output of the channel
decoder into a symbol sequence. The decoder for a system using fixed – length code
words is quite simple, but the decoder for a system using variable – length code words
will be very complex.

Aim of the source coding is to remove the redundancy in the transmitting

information, so that bandwidth required for transmission is minimized. Based on the
probability of the symbol code word is assigned. Higher the probability, shorter is the
codeword.
Ex: Huffman coding.

CHANNEL ENCODER / DECODER:

Error control is accomplished by the channel coding operation that consists of

systematically adding extra bits to the output of the source coder. These extra bits do not
convey any information but helps the receiver to detect and / or correct some of the errors
in the information bearing bits.

There are two methods of channel coding:

1. Block Coding: The encoder takes a block of ‘k’ information bits from the source
encoder and adds ‘r’ error control bits, where ‘r’ is dependent on ‘k’ and error
control capabilities desired.
2. Convolution Coding: The information bearing message stream is encoded in a
continuous fashion by continuously interleaving information bits and error control
bits.

The Channel decoder recovers the information bearing bits from the coded binary stream.
Error detection and possible correction is also performed by the channel decoder.
The important parameters of coder / decoder are: Method of coding, efficiency, error
control capabilities and complexity of the circuit.

MODULATOR:
The Modulator converts the input bit stream into an electrical waveform suitable
for transmission over the communication channel. Modulator can be effectively used to
minimize the effects of channel noise, to match the frequency spectrum of transmitted
signal with channel characteristics, to provide the capability to multiplex many signals.

DEMODULATOR:
The extraction of the message from the information bearing waveform produced
by the modulation is accomplished by the demodulator. The output of the demodulator is
bit stream. The important parameter is the method of demodulation.
CHANNEL:
The Channel provides the electrical connection between the source and
destination. The different channels are: Pair of wires, Coaxial cable, Optical fibre, Radio
channel, Satellite channel or combination of any of these.
The communication channels have only finite Bandwidth, non-ideal frequency
response, the signal often suffers amplitude and phase distortion as it travels over the
channel. Also, the signal power decreases due to the attenuation of the channel. The
signal is corrupted by unwanted, unpredictable electrical signals referred to as noise.
The important parameters of the channel are Signal to Noise power Ratio (SNR),
usable bandwidth, amplitude and phase response and the statistical properties of noise.

Modified Block Diagram: (With additional blocks)

From Other Sources

Source Encrypt Channel Modula

Source Encoder er Encoder Mux tor

Channel

Destina De Demod
Source
tion Decoder cryptor Channel Demux ulator
decoder

To other Destinations

Fig 1.3 : Block diagram with additional blocks

Some additional blocks as shown in the block diagram are used in most of digital
communication system:

 Encryptor: Encryptor prevents unauthorized users from understanding the

messages and from injecting false messages into the system.

 MUX : Multiplexer is used for combining signals from different sources so that
they share a portion of the communication system.
  DeMUX: DeMultiplexer is used for separating the different signals so that they
reach their respective destinations.

 Decryptor: It does the reverse operation of that of the Encryptor.

Synchronization: Synchronization involves the estimation of both time and frequency

coherent systems need to synchronize their frequency reference with carrier in both
frequency and phase.

Advantages of Digital Communication

1. The effect of distortion, noise and interference is less in a digital communication

system. This is because the disturbance must be large enough to change the pulse
from one state to the other.

2. Regenerative repeaters can be used at fixed distance along the link, to identify and
regenerate a pulse before it is degraded to an ambiguous state.

3. Digital circuits are more reliable and cheaper compared to analog circuits.

4. The Hardware implementation is more flexible than analog hardware because of

the use of microprocessors, VLSI chips etc.

5. Signal processing functions like encryption, compression can be employed to

maintain the secrecy of the information.

6. Error detecting and Error correcting codes improve the system performance by
reducing the probability of error.

7. Combining digital signals using TDM is simpler than combining analog signals
using FDM. The different types of signals such as data, telephone, TV can be
treated as identical signals in transmission and switching in a digital
communication system.

8. We can avoid signal jamming using spread spectrum technique.

Disadvantages of Digital Communication:

1. Large System Bandwidth:- Digital transmission requires a large system

bandwidth to communicate the same information in a digital format as compared
to analog format.

2. System Synchronization:- Digital detection requires system synchronization

whereas the analog signals generally have no such requirement.
Channels for Digital Communications

The modulation and coding used in a digital communication system depend on the
characteristics of the channel. The two main characteristics of the channel are
BANDWIDTH and POWER. In addition the other characteristics are whether the channel
is linear or nonlinear, and how free the channel is free from the external interference.

Five channels are considered in the digital communication, namely: telephone

channels, coaxial cables, optical fibers, microwave radio, and satellite channels.

Telephone channel: It is designed to provide voice grade communication. Also good for
data communication over long distances. The channel has a band-pass characteristic
occupying the frequency range 300Hz to 3400hz, a high SNR of about 30db, and
approximately linear response.
For the transmission of voice signals the channel provides flat amplitude
response. But for the transmission of data and image transmissions, since the phase delay
variations are important an equalizer is used to maintain the flat amplitude response and a
linear phase response over the required frequency band. Transmission rates upto16.8
kilobits per second have been achieved over the telephone lines.

Coaxial Cable: The coaxial cable consists of a single wire conductor centered inside an
outer conductor, which is insulated from each other by a dielectric. The main advantages
of the coaxial cable are wide bandwidth and low external interference. But closely
spaced repeaters are required. With repeaters spaced at 1km intervals the data rates of
274 megabits per second have been achieved.

Optical Fibers: An optical fiber consists of a very fine inner core made of silica glass,
surrounded by a concentric layer called cladding that is also made of glass. The refractive
index of the glass in the core is slightly higher than refractive index of the glass in the
cladding. Hence if a ray of light is launched into an optical fiber at the right oblique
acceptance angle, it is continually refracted into the core by the cladding. That means the
difference between the refractive indices of the core and cladding helps guide the
propagation of the ray of light inside the core of the fiber from one end to the other.
Compared to coaxial cables, optical fibers are smaller in size and they offer higher
transmission bandwidths and longer repeater separations.

Microwave radio: A microwave radio, operating on the line-of-sight link, consists

basically of a transmitter and a receiver that are equipped with antennas. The antennas
are placed on towers at sufficient height to have the transmitter and receiver in line-of-
sight of each other. The operating frequencies range from 1 to 30 GHz.
Under normal atmospheric conditions, a microwave radio channel is very reliable
and provides path for high-speed digital transmission. But during meteorological
variations, a severe degradation occurs in the system performance.
Satellite Channel: A Satellite channel consists of a satellite in geostationary orbit, an
uplink from ground station, and a down link to another ground station. Both link operate
at microwave frequencies, with uplink the uplink frequency higher than the down link
frequency. In general, Satellite can be viewed as repeater in the sky. It permits
communication over long distances at higher bandwidths and relatively low cost.

Chapter-2
SAMPLING PROCESS

SAMPLING: A message signal may originate from a digital or analog source. If

the message signal is analog in nature, then it has to be converted into digital form before
it can transmitted by digital means. The process by which the continuous-time signal is
converted into a discrete–time signal is called Sampling.
Sampling operation is performed in accordance with the sampling theorem.

SAMPLING THEOREM FOR LOW-PASS SIGNALS:-

Statement:- “If a band –limited signal g(t) contains no frequency components for ‫׀‬f‫ > ׀‬W,
then it is completely described by instantaneous values g(kTs) uniformly spaced in time
with period Ts ≤ 1/2W. If the sampling rate, fs is equal to the Nyquist rate or greater (fs ≥
2W), the signal g(t) can be exactly reconstructed.
g(t)

sδ (t)

-2Ts -Ts 0 1Ts 2Ts 3Ts 4Ts

gδ(t)

-2Ts -Ts 0 Ts 2Ts 3Ts 4Ts

Fig 2.1: Sampling process
Proof:- Consider the signal g(t) is sampled by using a train of impulses sδ (t).
Let gδ(t) denote the ideally sampled signal, can be represented as
gδ(t) = g(t).sδ(t) ------------------- 2.1
where sδ(t) – impulse train defined by

sδ(t) =   (t  kT
k  
s ) -------------------- 2.2


Therefore gδ(t) = g(t) .   (t  kT
k  
s )


=  g (kT ). (t  kT )
k  
s s ----------- 2.3

The Fourier transform of an impulse train is given by


Sδ(f )= F[sδ(t)] = fs  ( f
n  
 nf s ) ------------------ 2.4

Applying F.T to equation 2.1 and using convolution in frequency domain property,
Gδ(f) = G(f) * Sδ (f)

Using equation 2.4, Gδ (f) = G(f) * fs  ( f
n  
 nf s )


Gδ (f) = fs  G( f
n  
 nf s ) ----------------- 2.5

Fig. 2.2 Over Sampling (fs > 2W)

Fig. 2.3 Nyquist Rate Sampling (fs = 2W)

Fig. 2.4 Under Sampling (fs < 2W)

Reconstruction of g(t) from g δ (t):
By passing the ideally sampled signal gδ(t) through an low pass filter ( called
Reconstruction filter ) having the transfer function H R(f) with bandwidth, B satisfying
the condition W ≤ B ≤ (fs – W) , we can reconstruct the signal g(t). For an ideal
reconstruction filter the bandwidth B is equal to W.

Reconstruction
gδ (t) Filter gR(t)
HR(f) / hR(t)

The output of LPF is, gR(t) = gδ (t) * hR(t)

where hR(t) is the impulse response of the filter.
In frequency domain, GR(f) = Gδ(f) .HR(f).
For the ideal LPF HR(f) = K -W ≤ f ≤ +W
0 otherwise
then impulse response is hR(t) = 2WTs. Sinc(2Wt)
Correspondingly the reconstructed signal is
gR(t) = [ 2WTs Sinc (2Wt)] * [g δ (t)]

gR(t) = 2WTs  g (kTs ).Sinc(2Wt ) *  (t  kTs )
K  


gR(t) = 2WTs  g (kTs ).Sinc[2W (t  kTs )]
K  

Gδ(f)

-fs -W 0 W fs f
HR( f) K
-W 0 +W f
GR(f)
f
-W 0 +W
Fig: 2.5 Spectrum of sampled signal and reconstructed signal
Sampling of Band Pass Signals:
 Consider a band-pass signal g(t) with the spectrum shown in figure 2.6:
G(f)
B B
Band width = B
Upper Limit = fu
Lower Limit = fl -fu -fl 0 fl fu f
Fig 2.6: Spectrum of a Band-pass Signal

The signal g(t) can be represented by instantaneous values, g(kTs) if the sampling
rate fs is (2fu/m) where m is an integer defined as
((fu / B) -1 ) < m ≤ (fu / B)
If the sample values are represented by impulses, then g(t) can be exactly
reproduced from it’s samples by an ideal Band-Pass filter with the response, H(f) defined
as
H(f) = 1 fl < | f | <fu
0 elsewhere
If the sampling rate, fs ≥ 2fu, exact reconstruction is possible in which case the signal
g(t) may be considered as a low pass signal itself.

fs

4B
3B
2B

0 B 2B 3B 4B 5B fu

Fig 2.7: Relation between Sampling rate, Upper cutoff frequency and Bandwidth.
Example-1 :
Consider a signal g(t) having the Upper Cutoff frequency, f u = 100KHz and the
Lower Cutoff frequency fl = 80KHz.
The ratio of upper cutoff frequency to bandwidth of the signal g(t) is
fu / B = 100K / 20K = 5.
Therefore we can choose m = 5.
Then the sampling rate is fs = 2fu / m = 200K / 5 = 40KHz

Example-2 :
Consider a signal g(t) having the Upper Cutoff frequency, f u = 120KHz and the
Lower Cutoff frequency fl = 70KHz.
The ratio of upper cutoff frequency to bandwidth of the signal g(t) is
fu / B = 120K / 50K = 2.4
Therefore we can choose m = 2. ie.. m is an integer less than (fu /B).
Then the sampling rate is fs = 2fu / m = 240K / 2 = 120KHz
Quadrature Sampling of Band – Pass Signals:
This scheme represents a natural extension of the sampling of low – pass signals.
In this scheme, the band pass signal is split into two components, one is in-phase
component and other is quadrature component. These two components will be low–pass
signals and are sampled separately. This form of sampling is called quadrature sampling.
Let g(t) be a band pass signal, of bandwidth ‘2W’ centered around the frequency,
fc, (fc>W). The in-phase component, gI(t) is obtained by multiplying g(t) with
cos(2πfct) and then filtering out the high frequency components. Parallelly a quadrature
phase component is obtained by multiplying g(t) with sin(2πfct) and then filtering out the
high frequency components..
The band pass signal g(t) can be expressed as,
g(t) = gI(t). cos(2πfct) – gQ(t) sin(2πfct)
The in-phase, gI(t) and quadrature phase gQ(t) signals are low–pass signals, having band
limited to (-W < f < W). Accordingly each component may be sampled at the rate of
2W samples per second.

g(t)cos(2πfct) ½ gI (t) ½ gI (nTs)

LPF sampler
g(t)
cos (2πfct)
g(t) sin(2πfct) ½gQ(t) -½ gQ(nTs)
LPF

sampler

sin (2πfct)

Fig 2.8: Generation of in-phase and quadrature phase samples

 G(f)

-fc 0 fc f
 2W->
a) Spectrum of a Band pass signal.

GI(f) / GQ(f)

-W 0 W f
b) Spectrum of gI(t) and gQ(t)
Fig 2.9 a) Spectrum of Band-pass signal g(t)
b) Spectrum of in-phase and quadrature phase signals

RECONSTRUCTION:
From the sampled signals gI(nTs) and gQ(nTs), the signals gI(t) and gQ(t) are
obtained. To reconstruct the original band pass signal, multiply the signals g I(t) by
cos(2πfct) and sin(2πfct) respectively and then add the results.
gI(nTs)
Reconstruction
Filter

+
Cos (2πfct) Σ g(t)
-

gQ(nTs) Reconstruction
Filter

Sin (2πfct)
Fig 2.10: Reconstruction of Band-pass signal g(t)
Sample and Hold Circuit for Signal Recovery.

In both the natural sampling and flat-top sampling methods, the spectrum of the signals
are scaled by the ratio τ/Ts, where τ is the pulse duration and Ts is the sampling period.
Since this ratio is very small, the signal power at the output of the reconstruction filter is
correspondingly small. To overcome this problem a sample-and-hold circuit is used .

SW
AMPLIFIER

Input Output
g(t) u(t)

a) Sample and Hold Circuit

b) Idealized output waveform of the circuit

Fig: 2.17 Sample Hold Circuit with Waveforms.

The Sample-and-Hold circuit consists of an amplifier of unity gain and low output
impedance, a switch and a capacitor; it is assumed that the load impedance is large. The
switch is timed to close only for the small duration of each sampling pulse, during which
time the capacitor charges up to a voltage level equal to that of the input sample. When
the switch is open , the capacitor retains the voltage level until the next closure of the
switch. Thus the sample-and-hold circuit produces an output waveform that represents a
staircase interpolation of the original analog signal.
The output of a Sample-and-Hold circuit is defined as

u (t )   g (nTs ) h(t  nTs )
n  

where h(t) is the impulse response representing the action of the Sample-and-Hold
circuit; that is

h(t) = 1 for 0 < t < Ts

0 for t < 0 and t > Ts

Correspondingly, the spectrum for the output of the Sample-and-Hold circuit is given
by,

U ( f )  fs  H ( f )G ( f
n  
 nf s ) )

where G(f) is the FT of g(t) and

H(f) = Ts Sinc( fTs) exp( -jfTs)

To recover the original signal g(t) without distortion, the output of the Sample-and-
Hold circuit is passed through a low-pass filter and an equalizer.

Sampled Sample and Low Pass Analog

Waveform Hold Circuit Filter Equalizer Waveform

Fig. 2.18: Components of a scheme for signal reconstruction

Signal Distortion in Sampling.

In deriving the sampling theorem for a signal g(t) it is assumed that the signal g(t) is
strictly band-limited with no frequency components above ‘W’ Hz. However, a signal
cannot be finite in both time and frequency. Therefore the signal g(t) must have infinite
duration for its spectrum to be strictly band-limited.

In practice, we have to work with a finite segment of the signal in which case the
spectrum cannot be strictly band-limited. Consequently when a signal of finite duration
is sampled an error in the reconstruction occurs as a result of the sampling process.

Consider a signal g(t) whose spectrum G(f) decreases with the increasing frequency
without limit as shown in the figure 2.19. The spectrum, G (f) of the ideally sampled
signal , g(t) is the sum of G(f) and infinite number of frequency shifted replicas of
G(f). The replicas of G(f) are shifted in frequency by multiples of sampling frequency,
fs. Two replicas of G(f) are shown in the figure 2.19.

The use of a low-pass reconstruction filter with it’s pass band extending from (-fs/2 to
+fs/2) no longer yields an undistorted version of the original signal g(t). The portions of
the frequency shifted replicas are folded over inside the desired spectrum. Specifically,
high frequencies in G(f) are reflected into low frequencies in G(f). The phenomenon of
overlapping in the spectrum is called as Aliasing or Foldover Effect. Due to this
phenomenon the information is invariably lost.

Fig. 2.19 : a) Spectrum of finite energy signal g(t)

b) Spectrum of the ideally sampled signal.
Bound On Aliasing Error:

Let g(t) be the message signal, g(n/fs) denote the sequence obtained by sampling the
signal g(t) and gi(t) denote the signal reconstructed from this sequence by interpolation;
that is
 n 
g i (t )   g   Sinc( f s t  n)
n  fs 

Aliasing Error is given by,  = | g(t) - gi(t) |

Signal g(t) is given by


g (t )   G ( f ) exp( j 2ft ) df


Or equivalently

 ( m 1 / 2 ) fs

g (t )  
m  
 G ( f ) exp( j 2ft )df
( m 1 / 2 ) fs

Using Poisson’s formula and Fourier Series expansions we can obtain the aliasing error
as

 ( m 1 / 2 ) fs

  |  [1  exp( j 2mf s t )]
m  
 G ( f ) exp( j 2ft )df |
( m 1 / 2 ) fs

Correspondingly the following observations can be done :

1. The term corresponding to m=0 vanishes.

2. The absolute value of the sum of a set of terms is less than or equal to the sum of
the absolute values of the individual terms.
3. The absolute value of the term 1- exp(-j2πmfst) is less than or equal to 2.
4. The absolute value of the integral in the above equation is bounded as
( m 1 / 2 ) fs ( m 1 / 2 ) fs

|  G ( f ) exp( j 2ft )df

( m 1 / 2 ) fs
|  | G ( f ) | df
( m 1 / 2 ) fs

Hence the aliasing error is bounded as

  2 | f |  fs / 2
| G ( f ) | df
Example: Consider a time shifted sinc pulse, g(t) = 2 sinc(2t – 1). If g(t) is sampled at
rate of 1sample per second that is at t = 0, ± 1, ±2, ±3 and so on , evaluate
the aliasing error.

Solution: The given signal g(t) and it’s spectrum are shown in fig. 2.20.

2.0

1.0

t
-1 0 0.5 1 2

-1.0

a) Sinc Pulse

‫׀‬G(f)‫׀‬

-1.0 -1/2 0 1/2 1.0 f

(b) Amplitude Spectrum, ‫׀‬G(f)‫׀‬

Fig. 2.20

The sampled signal g(nTs) = 0 for n = 0, ± 1, ±2, ±3 . . . . .and reconstructed signal

gi(t) = 0 for all t.

From the figure, the sinc pulse attains it’s maximum value of 2 at time t equal to ½. The
aliasing error cannot exceed max|g(t)| = 2.
From the spectrum, the aliasing error is equal to unity.
Natural Sampling:
In this method of sampling, an electronic switch is used to periodically shift
between the two contacts at a rate of fs = (1/Ts ) Hz, staying on the input contact for C
seconds and on the grounded contact for the remainder of each sampling period.
The output xs(t) of the sampler consists of segments of x(t) and hence x s(t) can be
considered as the product of x(t) and sampling function s(t).
xs(t) = x(t) . s(t)
The sampling function s(t) is periodic with period Ts, can be defined as,
S(t) = 1  / 2 < t <  / 2 ------- (1)
0  / 2 < ‫׀‬t‫ < ׀‬Ts/2

Fig: 2.11 Natural Sampling – Simple Circuit.

Fig: 2.12 Natural Sampling – Waveforms.

 Using Fourier series, we can rewrite the signal S(t) as


S(t) = Co +  2Cn cos(nw t )

n 1
s

where the Fourier coefficients, Co =  / Ts & Cn = fs  Sinc(n fs  )


Therefore: xs(t) = x(t) [ Co +  2Cn cos(nw t ) ]
n 1
s

xs (t) = Co.x(t) +2C1.x(t)cos(wst) + 2C2.x(t)cos (2wst) + . . . . . . . .

Applying Fourier transform for the above equation
FT
Using x(t) X(f)
x(t) cos(2πf0t) ½ [X(f-f0) + X(f+f0)]

Xs(f) = Co.X(f) + C1 [X(f-f0) + X(f+f0)] + C2 [X(f-f0) + X(f+f0)] + ... …



Xs(f) = Co.X(f) + n  
Cn. X ( f  nfs )

n≠0

1 X(f)

f
-W 0 +W
Message Signal Spectrum
Xs(f)
C0
C2 C1 C1 C2

f
-2fs -fs -W 0 +W fs 2fs
Sampled Signal Spectrum (fs > 2W)

Fig:2.13 Natural Sampling Spectrum

The signal xs(t) has the spectrum which consists of message spectrum and repetition of
message spectrum periodically in the frequency domain with a period of f s. But the
message term is scaled by ‘Co”. Since the spectrum is not distorted it is possible to
reconstruct x(t) from the sampled waveform xs(t).

Flat Top Sampling:

In this method, the sampled waveform produced by practical sampling devices,
the pulse p(t) is a flat – topped pulse of duration, .

Fig. 2.14: Flat Top Sampling Circuit

Fig. 2.15: Waveforms

Mathematically we can consider the flat – top sampled signal as equivalent to the
convolved sequence of the pulse signal p(t) and the ideally sampled signal, x δ (t).
 xs(t) = p(t) *x δ (t)

xs(t) = p(t) * [  x(kTs ). (t - kTs) ]
k  

Applying F.T,
Xs(f) = P(f).X δ (f)

= P(f). fs  X(f
n  
 nfs )

where P(f) = FT[p(t)] and X δ (f) = FT[x δ (t)]

Aperature Effect:
The sampled signal in the flat top sampling has the attenuated high frequency
components. This effect is called the Aperture Effect.

The aperture effect can be compensated by:

1. Selecting the pulse width  as very small.
2. by using an equalizer circuit.

Sampled Signal
Low – Equalizer
Pass Filter Heq(f)

Equalizer decreases the effect of the in-band loss of the interpolation filter (lpf).
As the frequency increases, the gain of the equalizer increases. Ideally the amplitude
response of the equalizer is
1  f
| Heq(f)| = 1 / | P(f) | =  .SinC ( f )  Sin( f  )
 Chapter-3
Waveform Coding Techniques

PCM [Pulse Code Modulation]

PCM is an important method of analog –to-digital conversion. In this modulation

the analog signal is converted into an electrical waveform of two or more levels. A
simple two level waveform is shown in fig 3.1.

Fig:3.1 A simple binary PCM waveform

The PCM system block diagram is shown in fig 3.2. The essential operations in the
transmitter of a PCM system are Sampling, Quantizing and Coding. The Quantizing and
encoding operations are usually performed by the same circuit, normally referred to as
analog to digital converter.
The essential operations in the receiver are regeneration, decoding and
demodulation of the quantized samples. Regenerative repeaters are used to reconstruct
the transmitted sequence of coded pulses in order to combat the accumulated effects of
signal distortion and noise.

PCM Transmitter:

Basic Blocks:
1. Anti aliasing Filter
2. Sampler
3. Quantizer
4. Encoder

An anti-aliasing filter is basically a filter used to ensure that the input signal to sampler is
free from the unwanted frequency components.
For most of the applications these are low-pass filters. It removes the frequency
components of the signal which are above the cutoff frequency of the filter. The cutoff
frequency of the filter is chosen such it is very close to the highest frequency component
of the signal.
Sampler unit samples the input signal and these samples are then fed to the Quantizer
which outputs the quantized values for each of the samples. The quantizer output is fed
to an encoder which generates the binary code for every sample. The quantizer and
encoder together is called as analog to digital converter.

Continuous time
message signal PCM Wave

LPF Sampler Quantizer Encoder

(a) TRANSMITTER

Distorted Regenerative Regenerative

PCM wave Repeater Repeater

(b) Transmission Path

Input
Regeneration Decoder Reconstruction Destination
Circuit Filter User

(c) RECEIVER

Fig: 3.2 - PCM System : Basic Block Diagram

 REGENERATIVE REPEATER

REGENERATION: The feature of the PCM systems lies in the ability to control the
effects of distortion and noise produced by transmitting a PCM wave through a channel.
This is accomplished by reconstructing the PCM wave by means of regenerative
repeaters.
Three basic functions: Equalization
Timing and
Decision Making

Decision
Distorted Amplifier - Making Regenerated
PCM Equalizer Device PCM wave
Wave

Timing
Circuit

Fig: 3.3 - Block diagram of a regenerative repeater.

The equalizer shapes the received pulses so as to compensate for the effects of
amplitude and phase distortions produced by the transmission characteristics of the
channel.
The timing circuit provides a periodic pulse train, derived from the received
pulses, for sampling the equalized pulses at the instants of time where the signal to noise
ratio is maximum.
The decision device is enabled at the sampling times determined by the timing
circuit. It makes it’s decision based on whether the amplitude of the quantized pulse plus
noise exceeds a predetermined voltage level.
Quantization Process:

The process of transforming Sampled amplitude values of a message signal into a

discrete amplitude value is referred to as Quantization.

The quantization Process has a two-fold effect:

1. the peak-to-peak range of the input sample values is subdivided into a finite set
of decision levels or decision thresholds that are aligned with the risers of the
staircase, and
2. the output is assigned a discrete value selected from a finite set of representation
levels that are aligned with the treads of the staircase..

A quantizer is memory less in that the quantizer output is determined only by the value of
a corresponding input sample, independently of earlier analog samples applied to the
input.

Analog Signal

Discrete Samples
( Quantized )

0 Ts 2Ts 3Ts Time

Fig:3.4 Typical Quantization process.

Types of Quantizers:

1. Uniform Quantizer
2. Non- Uniform Quantizer
 In Uniform type, the quantization levels are uniformly spaced, whereas in non-
uniform type the spacing between the levels will be unequal and mostly the relation is
logarithmic.

Types of Uniform Quantizers: ( based on I/P - O/P Characteristics)

1. Mid-Rise type Quantizer
2. Mid-Tread type Quantizer

In the stair case like graph, the origin lies the middle of the tread portion in Mid –Tread
type where as the origin lies in the middle of the rise portion in the Mid-Rise type.

Mid – tread type: Quantization levels – odd number.

Mid – Rise type: Quantization levels – even number.

Output
7Δ/2

5Δ/2
3Δ/2

Δ/2

Δ 2Δ 3Δ 4Δ Input

Fig:3.5 Input-Output Characteristics of a Mid-Rise type Quantizer

 Output

2Δ

Δ/2
3Δ/2 Input

Fig:3.6 Input-Output Characteristics of a Mid-Tread type Quantizer

Quantization Noise and Signal-to-Noise:

“The Quantization process introduces an error defined as the difference between the input
signal, x(t) and the output signal, yt). This error is called the Quantization Noise.”

q(t) = x(t) – y(t)

Quantization noise is produced in the transmitter end of a PCM system by

rounding off sample values of an analog base-band signal to the nearest permissible
representation levels of the quantizer. As such quantization noise differs from channel
noise in that it is signal dependent.

Let ‘Δ’ be the step size of a quantizer and L be the total number of quantization levels.
Quantization levels are 0, ± Δ., ± 2 Δ., ±3 Δ . . . . . . .
The Quantization error, Q is a random variable and will have its sample values bounded
by [-(Δ/2) < q < (Δ/2)]. If Δ is small, the quantization error can be assumed to a
uniformly distributed random variable.

Consider a memory less quantizer that is both uniform and symmetric.

L = Number of quantization levels
X = Quantizer input
Y = Quantizer output
The output y is given by
Y=Q(x) ------------- (3.1)

which is a staircase function that befits the type of mid tread or mid riser quantizer of
interest.

Suppose that the input ‘x’ lies inside the interval

Ik = {xk < x ≤ xk+1} k = 1,2,---------L ------- ( 3.2)

where xk and xk+1 are decision thresholds of the interval Ik as shown in figure 3.7.

Ik-1 Ik

yk-1 yk
Xk-1 Xk Xk+1
Fig:3.7 Decision thresholds of the equalizer

Correspondingly, the quantizer output y takes on a discrete value

Y = yk if x lies in the interval Ik

Let q = quantization error with values in the range    q  
2 2 then
Yk = x+q if ‘n’ lies in the interval Ik
Assuming that the quantizer input ‘n’ is the sample value of a random variable ‘X’ of
zero mean with variance x 2 .

The quantization noise uniformly distributed through out the signal band, its interfering
effect on a signal is similar to that of thermal noise.

Expression for Quantization Noise and SNR in PCM:-

Let Q = Random Variable denotes the Quantization error

q = Sampled value of Q
Assuming that the random variable Q is uniformly distributed over the possible range
(-Δ/2 to Δ/2) , as

fQ(q) = 1/Δ - Δ/2 ≤ q ≤ Δ/2 ------- (3.3)

0 otherwise
 where fQ(q) = probability density function of the Quantization error. If the signal
does not overload the Quantizer, then the mean of Quantization error is zero and
its variance σQ2 .
fQ(q)
1/Δ

- Δ/2 0 Δ/2 q

Fig:3.8 PDF for Quantization error.

Therefore
2
 Q  E{Q 2 }

Q2  q

2
f q ( q ) dq ---- ( 3.4)

1 2 2
  q 2 dq 
2
Q --- (3.5)
  12

2

Thus the variance of the Quantization noise produced by a Uniform Quantizer,

grows as the square of the step size. Equation (3.5) gives an expression for Quantization
noise in PCM system.
2
Let  X = Variance of the base band signal x(t) at the input of Quantizer.
When the base band signal is reconstructed at the receiver output, we obtain
original signal plus Quantization noise. Therefore output signal to Quantization noise
ration (SNR) is given by

Signal Power  X  2 2

( SNR) O    2X ---------- (3.6)

Noise Power  Q 2
 / 12

Smaller the step size Δ, larger will be the SNR.

Signal to Quantization Noise Ratio:- [ Mid Tread Type ]

Let x = Quantizer input, sampled value of random variable X with mean X, variance
2
 X . The Quantizer is assumed to be uniform, symmetric and mid tread type.
xmax = absolute value of the overload level of the Quantizer.
Δ = Step size
L = No. of Quantization level given by
 2 x max
L 1 ----- (3.7)


Let n = No. of bits used to represent each level.

In general 2n = L, but in the mid tread Quantizer, since the number of representation
levels is odd,

L = 2n – 1 --------- (Mid tread only) ---- (3.8)

From the equations 3.7 and 3.8,

2 x max
2n 1  1

Or

x
 max
n 1 ---- (3.9)
2 1
x max
The ratio is called the loading factor. To avoid significant overload distortion, the
x
amplitude of the Quantizer input x extend from  4 x to 4 x , which corresponds to
loading factor of 4. Thus with x max  4 x we can write equation (3.9) as

4 x
 n 1 ----------(3.10)
2 1
 X 2 3 n1
( SNR ) O  2  [2  1] 2 -------------(3.11)
 / 12 4

For larger value of n (typically n>6), we may approximate the result as

3 n 1 3
( SNR) O  [2  1] 2  (2 2 n ) --------------- (3.12)
4 16
Hence expressing SNR in db

10 log10 (SNR)O = 6n - 7.2 --------------- (3.13)

This formula states that each bit in codeword of a PCM system contributes 6db to the
signal to noise ratio.

For loading factor of 4, the problem of overload i.e. the problem that the sampled
value of signal falls outside the total amplitude range of Quantizer, 8σx is less than 10-4.

The equation 3.11 gives a good description of the noise performance of a PCM
system provided that the following conditions are satisfied.
1. The Quantization error is uniformly distributed
 2. The system operates with an average signal power above the error threshold so
that the effect of channel noise is made negligible and performance is there by
limited essentially by Quantization noise alone.
3. The Quantization is fine enough (say n>6) to prevent signal correlated patterns in
the Quantization error waveform
4. The Quantizer is aligned with input for a loading factor of 4

Note: 1. Error uniformly distributed

2. Average signal power
3. n > 6
4. Loading factor = 4

From (3.13): 10 log10 (SNR)O = 6n – 7.2

In a PCM system, Bandwidth B = nW or [n=B/W]

substituting the value of ‘n’ we get
10 log10 (SNR)O = 6(B/W) – 7.2 --------(3.14)

Signal to Quantization Noise Ratio:- [ Mid Rise Type ]

2
Let x = Quantizer input, sampled value of random variable X with mean X variance  X
.
The Quantizer is assumed to be uniform, symmetric and mid rise type.
Let xmax = absolute value of the overload level of the Quantizer.

2x max
L  ------------------(3.15)


Since the number of representation levels is even,

L = 2n ------- (Mid rise only) ---- (3.16)

From (3.15) and (3.16)

xmax
 -------------- (3.17)
2n

2

( SNR ) O  2 X -------------(3.18)
 / 12

where  X
2
represents the variance or the signal power.
Consider a special case of Sinusoidal signals:

Let the signal power be Ps, then Ps = 0.5 x2max.

Ps 12 Ps
( SNR ) O    1.5 L2  1.5 2 2 n -----(3.19)
 / 12
2
2

In decibels, ( SNR )0 = 1.76 + 6.02 n -----------(3.20)

Improvement of SNR can be achieved by increasing the number of bits, n. Thus

for ‘n’ number of bits / sample the SNR is given by the above equation 3.19. For every
increase of one bit / sample the step size reduces by half. Thus for (n+1) bits the SNR is
given by
(SNR) (n+1) bit = (SNR) (n) bit + 6dB

Therefore addition of each bit increases the SNR by 6dB

Problem-1: An analog signal is sampled at the Nyquist rate fs = 20K and quantized
into L=1024 levels. Find Bit-rate and the time duration Tb of one bit of the binary
encoded signal.

Solution: Assume Mid-rise type, n = log2L = 10

Bit-rate = Rb = nfs = 200K bits/sec
Bit duration Tb = 1/ Rb = 5µsec.

Problem-2: A PCM system uses a uniform quantizer followed by a 7-bit binary encoder.
The bit rate of the system is 56Mega bits/sec. Find the output signal-to-quantization
noise ratio when a sinusoidal wave of 1MHz frequency is applied to the input.

Solution:
Given n = 7 and bit rate Rb = 56 Mega bits per second.
Sampling frequency = Rb/n = 8MHz
Message bandwidth = 4MHz.
For Mid-rise type
(SNR)0 = 43.9 dB
CLASSIFICATION OF QUANTIZATION NOISE:
The Quantizing noise at the output of the PCM decoder can be categorized into
four types depending on the operating conditions:
Overload noise, Random noise, Granular Noise and Hunting noise

OVER LOAD NOISE:- The level of the analog waveform at the input of the PCM
encoder needs to be set so that its peak value does not exceed the design peak of Vmax
volts. If the peak input does exceed Vmax, then the recovered analog waveform at the
output of the PCM system will have flat – top near the peak values. This produces
overload noise.

GRANULAR NOISE:- If the input level is reduced to a relatively small value w.r.t to the
design level (quantization level), the error values are not same from sample to sample and
the noise has a harsh sound resembling gravel being poured into a barrel. This is granular
noise.
This noise can be randomized (noise power decreased) by increasing the number
of quantization levels i.e.. increasing the PCM bit rate.

HUNTING NOISE:- This occurs when the input analog waveform is nearly constant.
For these conditions, the sample values at the Quantizer output can oscillate between two
adjacent quantization levels, causing an undesired sinusoidal type tone of frequency
(0.5fs) at the output of the PCM system
This noise can be reduced by designing the quantizer so that there is no vertical
step at constant value of the inputs.
ROBUST QUANTIZATION

Features of an uniform Quantizer

– Variance is valid only if the input signal does not overload Quantizer
– SNR Decreases with a decrease in the input power level.

A Quantizer whose SNR remains essentially constant for a wide range of input power
levels. A quantizer that satisfies this requirement is said to be robust. The provision for
such robust performance necessitates the use of a non-uniform quantizer. In a non-
uniform quantizer the step size varies. For smaller amplitude ranges the step size is small
and larger amplitude ranges the step size is large.

In Non – Uniform Quantizer the step size varies. The use of a non – uniform
quantizer is equivalent to passing the baseband signal through a compressor and then
applying the compressed signal to a uniform quantizer. The resultant signal is then
transmitted.

UNIFORM
COMPRESSOR QUANTIZER EXPANDER

Fig: 3.9 MODEL OF NON UNIFORM QUANTIZER

At the receiver, a device with a characteristic complementary to the compressor

called Expander is used to restore the signal samples to their correct relative level.

The Compressor and expander take together constitute a Compander.

Compander = Compressor + Expander

Advantages of Non – Uniform Quantization :

1. Higher average signal to quantization noise power ratio than the uniform quantizer
when the signal pdf is non uniform which is the case in many practical situation.
2. RMS value of the quantizer noise power of a non – uniform quantizer is
substantially proportional to the sampled value and hence the effect of the
quantizer noise is reduced.
Expression for quantization error in non-uniform quantizer:

The Transfer Characteristics of the compressor and expander are denoted by C(x)
and C-1(x) respectively, which are related by,

C(x). C-1(x) = 1 ------ ( 3.21 )

The Compressor Characteristics for large L and x inside the interval Ik:
dc ( x) 2 xmax
 for k  0,1,.....L  1 ---------- ( 3.22 )
dx L k
where Δk = Width in the interval Ik.

Let fX(x) be the PDF of ‘X’ .

Consider the two assumptions:

– fX(x) is Symmetric
– fX(x) is approximately constant in each interval. ie.. fX(x) = fX(yk)

Δk = xk+1 - xk for k = 0, 1, … L-1. ------(3.23)

Let pk = Probability of variable X lies in the interval Ik, then

pk = P (xk < X < xk+1) = fX(x) Δk = fX(yk) Δk ------ (3.24)

L 1
with the constraint p
k 0
k 1

Let the random variable Q denote the quantization error, then

Q = yk – X for xk < X < xk+1

Variance of Q is
σQ2 = E ( Q2) = E [( X – yk )2 ] ---- (3.25)

 x max

 Q2   (x  y
 xmax
k ) 2 f X ( x) dx ---- ( 3.26)

Dividing the region of integration into L intervals and using (3.24)

L 1 x
pk k 1
Q 
2
 ( x  yk ) 2 dx ----- (3.27)
k 0  k xk

Using yk = 0.5 ( xk + xk+1 ) in 3.27 and carrying out the integration w.r.t x, we obtain
that
 1 L 1
Q   pk 2k
2
------- (3.28)
12 k 0

Compression Laws.

Two Commonly used logarithmic compression laws are called µ - law and A – law.

μ-law:
In this companding, the compressor characteristics is defined by equation 3.29.
The normalized form of compressor characteristics is shown in the figure 3.10. The μ-
law is used for PCM telephone systems in the USA, Canada and Japan. A practical
value for μ is 255.

c( x ) ln(1   x / x max ) x
 0 1
xmax ln(1   ) xmax
----( 3.29)

Fig: 3.10 Compression characteristics of μ-law

A-law:
In A-law companding the compressor characteristics is defined by equation 3.30. The
normalized form of A-law compressor characteristics is shown in the figure 3.11. The
A-law is used for PCM telephone systems in Europe. A practical value for A is 100.

A x / xmax x 1
0 
c( x ) 1  ln A xmax A

xmax
1  ln A x / x ma ) 1 x
 1
(1 l A x A x ma
n x
------------- ( 3.30)

Fig. 3.11: A-law compression Characteristics.

Advantages of Non Uniform Quantizer

• Reduced Quantization noise

• High average SNR

Differential Pulse Code Modulation (DPCM)

For the signals which does not change rapidly from one sample to next sample, the PCM
scheme is not preferred. When such highly correlated samples are encoded the resulting
encoded signal contains redundant information. By removing this redundancy before
encoding an efficient coded signal can be obtained. One of such scheme is the DPCM
technique. By knowing the past behavior of a signal up to a certain point in time, it is
possible to make some inference about the future values.

The transmitter and receiver of the DPCM scheme is shown in the fig3.12 and fig 3.13
respectively.

Transmitter: Let x(t) be the signal to be sampled and x(nTs) be it’s samples. In this
scheme the input to the quantizer is a signal

e(nTs) = x(nTs) - x^(nTs) ----- (3.31)

where x^(nTs) is the prediction for unquantized sample x(nTs). This predicted value is
produced by using a predictor whose input, consists of a quantized versions of the input
signal x(nTs). The signal e(nTs) is called the prediction error.

By encoding the quantizer output, in this method, we obtain a modified version of the
PCM called differential pulse code modulation (DPCM).

Quantizer output, v(nTs) = Q[e(nTs)]

= e(nTs) + q(nTs) ---- (3.32)
where q(nTs) is the quantization error.

Predictor input is the sum of quantizer output and predictor output,

u(nTs) = x^(nTs) + v(nTs) ---- (3.33)

Using 3.32 in 3.33, u(nTs) = x^(nTs) + e(nTs) + q(nTs) ----(3.34)

u(nTs) = x(nTs) + q(nTs) ----(3.35)

The receiver consists of a decoder to reconstruct the quantized error signal. The quantized
version of the original input is reconstructed from the decoder output using the same
predictor as used in the transmitter. In the absence of noise the encoded signal at the
receiver input is identical to the encoded signal at the transmitter output. Correspondingly
the receive output is equal to u(nTs), which differs from the input x(nts) only by the
quantizing error q(nTs).
 Sampled Input
x(nTs) e(nTs) v(nTs) Output
Σ Quantizer
+

^
x(nTs) Σ

Predictor
u(nTs)

Fig:3.12 - Block diagram of DPCM Transmitter

Input v(nTs) u(nTs)

Decoder Σ
b(nTs)
Output

x^(nTs)
Predictor

Fig:3.13 - Block diagram of DPCM Receiver.

Prediction Gain  ( Gp):

The output signal-to-quantization noise ratio of a signal coder is defined as

2
(SNR )0  X2 --------------------( 3.36)
Q
where σx2 is the variance of the signal x(nTs) and σQ2 is the variance of the
quantization error q(nTs). Then
 2   2 
(SNR ) 0   X2   E2   G P ( SNR ) P ------(3.37)
 
  E Q 
where σE2 is the variance of the prediction error e(nTs) and (SNR)P is the prediction
error-to-quantization noise ratio, defined by
 E2
(SNR) P  2 --------------(3.38)
Q
The Prediction gain Gp is defined as
 X2
GP  --------(3.39)
 E2
The prediction gain is maximized by minimizing the variance of the prediction error.
Hence the main objective of the predictor design is to minimize the variance of the
prediction error.
1
The prediction gain is defined by GP  ---- (3.40)
(1  12 )
and  E2   X2 (1  12 ) ----(3.41)
where ρ1 – Autocorrelation function of the message signal

PROBLEM:
Consider a DPCM system whose transmitter uses a first-order predictor optimized
in the minimum mean-square sense. Calculate the prediction gain of the system
for the following values of correlation coefficient for the message signal:
R (1) R (1)
(i ) 1  x  0.825 (ii ) 1  x  0.950
Rx (0) Rx (0)
Solution:
Using (3.40)
(i) For ρ1= 0.825, Gp = 3.13 In dB , Gp = 5dB

(ii) For ρ2 = 0.95, Gp = 10.26 In dB, Gp = 10.1dB

Delta Modulation (DM)

Delta Modulation is a special case of DPCM. In DPCM scheme if the base band signal
is sampled at a rate much higher than the Nyquist rate purposely to increase the
correlation between adjacent samples of the signal, so as to permit the use of a simple
quantizing strategy for constructing the encoded signal, Delta modulation (DM) is
precisely such as scheme. Delta Modulation is the one-bit (or two-level) versions of
DPCM.

DM provides a staircase approximation to the over sampled version of an input base band
signal. The difference between the input and the approximation is quantized into only two
levels, namely, ±δ corresponding to positive and negative differences, respectively, Thus,
if the approximation falls below the signal at any sampling epoch, it is increased by δ.
Provided that the signal does not change too rapidly from sample to sample, we find that
the stair case approximation remains within ±δ of the input signal. The symbol δ denotes
the absolute value of the two representation levels of the one-bit quantizer used in the
DM. These two levels are indicated in the transfer characteristic of Fig 3.14. The step
size  of the quantizer is related to δ by
 = 2δ ----- (3.42)

Output

+δ

0 Input

-δ

Fig-3.14: Input-Output characteristics of the delta modulator.

Let the input signal be x(t) and the staircase approximation to it is u(t). Then, the basic
principle of delta modulation may be formalized in the following set of relations:

e( nTs )  x ( nTs )  x ^ ( nTs )

e( nTs )  x ( nTs )  u ( nTs  Ts )
b ( nTs )   sgn[ e ( nTs )] and
u ( nTs )  u ( nTs  Ts )  b ( nTs )
----- (3.43)

where Ts is the sampling period; e(nTs) is a prediction error representing the difference
between the present sample value x(nTs) of the input signal and the latest approximation
 ^
to it, namely x(nT )  u (nT  T ) .The binary quantity, b(nTs ) is the one-bit word
s s s
transmitted by the DM system.

The transmitter of DM system is shown in the figure3.15. It consists of a summer, a two-

level quantizer, and an accumulator. Then, from the equations of (3.43) we obtain the
output as,
n n
u ( nTs )    sgn[e(iTs )]   b(iTs )
i 1 i 1
----- (3.44)
At each sampling instant, the accumulator increments the approximation to the input
signal by ±δ, depending on the binary output of the modulator.

Sampled Input
x(nTs) e(nTs) b(nTs) Output
Σ One - Bit
+ Quantizer

^
x(nTs) Σ

Delay
Ts u(nTs)

Fig 3.15 - Block diagram for Transmitter of a DM system

In the receiver, shown in fig.3.16, the stair case approximation u(t) is reconstructed by
passing the incoming sequence of positive and negative pulses through an accumulator in
a manner similar to that used in the transmitter. The out-of –band quantization noise in
the high frequency staircase waveform u(t) is rejected by passing it through a low-pass
filter with a band-width equal to the original signal bandwidth.

Delta modulation offers two unique features:

1. No need for Word Framing because of one-bit code word.

2. Simple design for both Transmitter and Receiver
 u(nTs) Low pass
Input Σ
Filter
b(nTs)

Delay
Ts
u(nTs-Ts)

Fig 3.16 - Block diagram for Receiver of a DM system

QUANTIZATION NOISE

Delta modulation systems are subject to two types of quantization error:

(1) slope –overload distortion, and (2) granular noise.

If we consider the maximum slope of the original input waveform x(t), it is clear that in
order for the sequence of samples{u(nTs)} to increase as fast as the input sequence of
samples {x(nTs)} in a region of maximum slope of x(t), we require that the condition in
equation 3.45 be satisfied.
 dx (t )
 max ------- ( 3.45 )
Ts dt
Otherwise, we find that the step size  = 2δ is too small for the stair case
approximation u(t) to follow a steep segment of the input waveform x(t), with the result
that u(t) falls behind x(t). This condition is called slope-overload, and the resulting
quantization error is called slope-overload distortion(noise). Since the maximum slope of
the staircase approximation u(t) is fixed by the step size  , increases and decreases in
u(t) tend to occur along straight lines. For this reason, a delta modulator using a fixed step
size is often referred ton as linear delta modulation (LDM).

The granular noise occurs when the step size  is too large relative to the local
slope characteristics of the input wave form x(t), thereby causing the staircase
approximation u(t) to hunt around a relatively flat segment of the input waveform; The
granular noise is analogous to quantization noise in a PCM system.

The e choice of the optimum step size that minimizes the mean-square value of
the quantizing error in a linear delta modulator will be the result of a compromise
between slope overload distortion and granular noise.

Output SNR for Sinusoidal Modulation.

Consider the sinusoidal signal, x(t) = A cos(2µfot)
The maximum slope of the signal x(t) is given by

dx (t )
max  2 f 0 A ----- (3.46)
dt
The use of Eq.5.81 constrains the choice of step size  = 2δ, so as to avoid slope-
overload. In particular, it imposes the following condition on the value of δ:

 dx(t )
 max  2 f 0 A ----- (3. 47)
Ts dt
Hence for no slope overload error the condition is given by equations 3.48 and 3.49.


A ------ (3.48)
2 f 0Ts

  2 f 0 ATs ------ (3.49)

Hence, the maximum permissible value of the output signal power equals

A2 2
Pmax   ---- (3.50)
2 8 2 f 02 Ts2
When there is no slope-overload, the maximum quantization error ±δ. Assuming that the
quantizing error is uniformly distributed (which is a reasonable approximation for small
δ). Considering the probability density function of the quantization error,( defined in
equation 3.51 ),

1
f Q (q )  for    q   
2 ----- (3.51)
0 otherwise
The variance of the quantization error is  2 Q .

1 2
Q q
2
 2
dq  ----- (3.52)
2 
3

The receiver contains (at its output end) a low-pass filter whose bandwidth is set equal to
the message bandwidth (i.e., highest possible frequency component of the message
signal), denoted as W such that f0 ≤ W. Assuming that the average power of the
quantization error is uniformly distributed over a frequency interval extending from -1/T s
to 1/Ts, we get the result:
 fc   2  2 
Average output noise power N o     WTs   ----- ( 3.53)
 fs  3  3 
Correspondingly, the maximum value of the output signal-to-noise ratio equals
 Pmax 3
(SNR)O   ----- (3.54)
No 8 Wf 02 Ts3
2

Equation 3.54 shows that, under the assumption of no slope-overload distortion, the
maximum output signal-to-noise ratio of a delta modulator is proportional to the sampling
rate cubed. This indicates a 9db improvement with doubling of the sampling rate.

Problems

1. Determine the output SNR in a DM system for a 1KHz sinusoid sampled

at 32KHz without slope overload and followed by a 4KHz post
reconstruction filter.

Solution:

Given W=4KHz, f0 = 1KHz , fs = 32KHz

Using equation (3.54) we get

(SNR)0 = 311.3 or 24.9dB

Delta Modulation:

Problems

2. Consider a Speech Signal with maximum frequency of 3.4KHz and

maximum amplitude of 1volt. This speech signal is applied to a delta modulator
whose bit rate is set at 60kbit/sec. Explain the choice of an appropriate step size for
the modulator.

Solution: Bandwidth of the signal = 3.4 KHz.

Maximum amplitude = 1 volt
Bit Rate = 60Kbits/sec
Sampling rate = 60K Samples/sec.
STEP SIZE = 0.356 Volts

3. Consider a Speech Signal with maximum frequency of 3.4KHz and

maximum amplitude of 1volt. This speech signal is applied to a delta modulator
whose bit rate is set at 20kbit/sec. Explain the choice of an appropriate step size for
the modulator.

Solution: Bandwidth of the signal = 3.4 KHz.

 Maximum amplitude = 1 volt
Bit Rate = 20Kbits/sec
Sampling rate = 20K Samples/sec.
STEP SIZE = 1.068 Volts

4. Consider a Delta modulator system designed to operate at 4 times the Nyquist

rate for a signal with a 4KHz bandwidth. The step size of the quantizer is 400mV.
a) Find the maximum amplitude of a 1KHz input sinusoid for which the delta
modulator does not show slope overload.
b) Find post-filtered output SNR

Solution: Bandwidth of the signal = f0 =1 KHz.

Nyquist Rate = 8K samples/sec
Sampling Rate = 32K samples/sec.
Step Size = 400 mV
a) For 1KHz sinusoid, Amax = 2.037 volts.
b) Assuming LPF bandwidth = W= 4KHz
SNR = 311.2586 = 24.93 dB

Adaptive Delta Modulation:

The performance of a delta modulator can be improved significantly by making the step
size of the modulator assume a time-varying form. In particular, during a steep segment
of the input signal the step size is increased. Conversely, when the input signal is varying
slowly, the step size is reduced. In this way, the size is adapted to the level of the input
signal. The resulting method is called adaptive delta modulation (ADM).

There are several types of ADM, depending on the type of scheme used for adjusting the
step size. In this ADM, a discrete set of values is provided for the step size. Fig.3.17
shows the block diagram of the transmitter and receiver of an ADM System.

In practical implementations of the system, the step size

 ( nTs ) or 2 ( nTs )
is constrained to lie between minimum and maximum values.

The upper limit,  max , controls the amount of slope-overload distortion. The lower limit,
 min , controls the amount of idle channel noise. Inside these limits, the adaptation rule
for  ( nTs ) is expressed in the general form

δ(nTs) = g(nTs). δ(nTs – Ts) ------ (3.55)

where the time-varying multiplier g ( nTs ) depends on the present binary output b(nTs )
of the delta modulator and the M previous values
b( nTs  Ts ), ....... b( nTs  MTs ) .

This adaptation algorithm is called a constant factor ADM with one-bit memory,
where the term “one bit memory” refers to the explicit utilization of the single pervious
bit b(nTs  Ts ) because equation (3.55) can be written as,

g(nTs) = K if b(nTs) = b(nTs – Ts)

g(nTs) = K-1 if b(nTs) = b(nTs – Ts) ------ (3.56)

This algorithm of equation (3.56), with K=1.5 has been found to be well matched to
typically speech and image inputs alike, for a wide range of bit rates.

Figure: 3.17a) Block Diagram of ADM Transmitter.

Figure: 3.17 b): Block Diagram of ADM Receiver.
Coding Speech at Low Bit Rates:

The use of PCM at the standard rate of 64 kb/s demands a high channel
bandwidth for its transmission. But channel bandwidth is at a premium, in which case
there is a definite need for speech coding at low bit rates, while maintaining acceptable
fidelity or quality of reproduction. The fundamental limits on bit rate suggested by speech
perception and information theory show that high quality speech coding is possible at
rates considerably less that 64 kb/s (the rate may actually be as low as 2 kb/s).

For coding speech at low bit rates, a waveform coder of prescribed configuration
is optimized by exploiting both statistical characterization of speech waveforms and
properties of hearing. The design philosophy has two aims in mind:
1. To remove redundancies from the speech signal as far as possible.
2. To assign the available bits to code the non-redundant parts of the speech signal in
a perceptually efficient manner.

To reduce the bit rate from 64 kb/s (used in standard PCM) to 32, 16, 8 and 4
kb/s, the algorithms for redundancy removal and bit assignment become increasingly
more sophisticated.

There are two schemes for coding speech:

1. Adaptive Differential Pulse code Modulation (ADPCM) --- 32 kb/s
2. Adaptive Sub-band Coding.--- 16 kb/s

1. Adaptive Differential Pulse – Code Modulation

A digital coding scheme that uses both adaptive quantization and adaptive
prediction is called adaptive differential pulse code modulation (ADPCM).

The term “adaptive” means being responsive to changing level and spectrum of the input
speech signal. The variation of performance with speakers and speech material, together
with variations in signal level inherent in the speech communication process, make the
combined use of adaptive quantization and adaptive prediction necessary to achieve best
performance.

The term “adaptive quantization” refers to a quantizer that operates with a time-varying
step size  ( nTs ) , where Ts is the sampling period. The step size  ( nTs ) is varied so
as to match the variance  2 x of the input signal x (nTs ) . In particular, we write

Δ(nTs) = Φ. σ^x(nTs) ----- (3.57)

where Φ – Constant
σ^x(nTs) – estimate of the σx(nTs)

Thus the problem of adaptive quantization, according to (3.57) is one of estimating

 x (nTs ) continuously.
 ^
The computation of the estimate  x (nT ) in done by one of two ways:
s
1. Unquantized samples of the input signal are used to derive forward estimates of
 x (nTs ) - adaptive quantization with forward estimation (AQF)
2. Samples of the quantizer output are used to derive backward estimates of
 x (nTs ) - adaptive quantization with backward estimation (AQB)

The use of adaptive prediction in ADPCM is required because speech signals are
inherently nonstationary, a phenomenon that manifests itself in the fact that
autocorrection function and power spectral density of speech signals are time-varying
functions of their respective variables. This implies that the design of predictors for such
inputs should likewise be time-varying, that is, adaptive. As with adaptive quantization,
there are two schemes for performing adaptive prediction:
1. Adaptive prediction with forward estimation (APF), in which unquantized
samples of the input signal are used to derive estimates of the predictor
coefficients.
2. Adaptive prediction with backward estimation (APB), in which samples of the
quantizer output and the prediction error are used to derive estimates of the
prediction error are used to derive estimates of the predictor coefficients.

(2) Adaptive Sub-band Coding:

PCM and ADPCM are both time-domain coders in that the speech signal is
processed in the time-domain as a single full band signal. Adaptive sub-band coding
is a frequency domain coder, in which the speech signal is divided into a number of
sub-bands and each one is encoded separately. The coder is capable of digitizing
speech at a rate of 16 kb/s with a quality comparable to that of 64 kb/s PCM. To
accomplish this performance, it exploits the quasi-periodic nature of voiced speech
and a characteristic of the hearing mechanism known as noise masking.

Periodicity of voiced speech manifests itself in the fact that people speak with a
characteristic pitch frequency. This periodicity permits pitch prediction, and
therefore a further reduction in the level of the prediction error that requires
quantization, compared to differential pulse code modulation without pitch
prediction. The number of bits per sample that needs to be transmitted is thereby
greatly reduced, without a serious degradation in speech quality.

In adaptive sub band coding (ASBC), noise shaping is accomplished by adaptive

bit assignment. In particular, the number of bits used to encode each sub-band is
varied dynamically and shared with other sub-bands, such that the encoding accuracy
is always placed where it is needed in the frequency – domain characterization of the
signal. Indeed, sub-bands with little or no energy may not be encoded at all.
Applications
1. Hierarchy of Digital Multiplexers
2. Light wave Transmission Link

(1) Digital Multiplexers:

Digital Multiplexers are used to combine digitized voice and video signals as
well as digital data into one data stream.

The digitized voice signals, digitized facsimile and television signals and
computer outputs are of different rates but using multiplexers it combined into a single
data stream.

1 1
2 2
: Multiplex
High-Speed
:
Transmission DeMux
: er
line :
N N

Fig. 3.18: Conceptual diagram of Multiplexing and Demultiplexing.

Two Major groups of Digital Multiplexers:

1. To combine relatively Low-Speed Digital signals used for voice-grade channels.

Modems are required for the implementation of this scheme.
2. Operates at higher bit rates for communication carriers.

Basic Problems associated with Multiplexers:

1. Synchronization.
2. Multiplexed signal should include Framing.
3. Multiplexer Should be capable handling Small variations
Digital Hierarchy based on T1 carrier:

This was developed by Bell system. The T1 carrier is designed to operate at 1.544 mega
bits per second, the T2 at 6.312 megabits per second, the T3 at 44.736 megabits per
second, and the T4 at 274.176 mega bits per second. This system is made up of various
combinations of lower order T-carrier subsystems. This system is designed to
accommodate the transmission of voice signals, Picture phone service and television
signals by using PCM and digital signals from data terminal equipment. The structure is
shown in the figure 3.19.

Fig. 3.19: Digital hierarchy of a 24 channel system.

The T1 carrier system has been adopted in USA, Canada and Japan. It is designed to
accommodate 24 voice signals. The voice signals are filtered with low pass filter having
cutoff of 3400 Hz. The filtered signals are sampled at 8KHz. The µ-law Companding
technique is used with the constant μ = 255.

With the sampling rate of 8KHz, each frame of the multiplexed signal occupies a period
of 125μsec. It consists of 24 8-bit words plus a single bit that is added at the end of the
frame for the purpose of synchronization. Hence each frame consists of a total 193 bits.
Each frame is of duration 125μsec, correspondingly, the bit rate is 1.544 mega bits per
second.
Another type of practical system, that is used in Europe is 32 channel system which is
shown in the figure 3.20.

Fig 3.20: 32 channel TDM system

32 channel TDM Hierarchy:

In the first level 2.048 megabits/sec is obtained by multiplexing 32 voice channels.

4 frames of 32 channels = 128 PCM channels,
Data rate = 4 x 2.048 Mbit/s = 8.192 Mbit/s,
But due to the synchronization bits the data rate increases to 8.448Mbit/sec.

4 x 128 = 512 channels

Data rate = 4 x8.192 Mbit/s (+ signalling bits) = 34.368 Mbit/s
(2) Light Wave Transmission
Optical fiber wave guides are very useful as transmission medium. They have a
very low transmission losses and high bandwidths which is essential for high-speed
communications. Other advantages include small size, light weight and immunity to
electromagnetic interference.

The basic optical fiber link is shown in the figure 3.21. The binary data fed into the
transmitter input, which emits the pulses of optical power., with each pulse being on or
off in accordance with the input data. The choice of the light source determines the
optical signal power available for transmission.

Fig: 3.21- Optical fiber link.

The on-off light pulses produced by the transmitter are launched into the optical fiber
wave guide. During the course of the propagation the light pulse suffers loss or
attenuation that increases exponentially with the distance.

At the receiver the original input data are regenerated by performing three basic
operations which are :
1. Detection – the light pulses are converted back into pulses of electrical current.
2. Pulse Shaping and Timing - This involves amplification, filtering and
equalization of the electrical pulses, as well as the extraction of timing
information.
3. Decision Making: Depending the pulse received it should be decided that the
received pulse is on or off.

--END—
e-Notes by Prof. H.V.Kumaraswamy, RVCE, Bangalore

CHAPTER 5

Digital modulation techniques

Modulation is defined as the process by which some characteristics of a carrier is
varied in accordance with a modulating wave. In digital communications, the modulating
wave consists of binary data or an M-ary encoded version of it and the carrier is
sinusoidal wave.
Different Shift keying methods that are used in digital modulation techniques are
 Amplitude shift keying [ASK]
 Frequency shift keying [FSK]
 Phase shift keying [PSK]
Fig shows different modulations
1. ASK[Amplitude Shift Keying]:

In a binary ASK system symbol ‘1’ and ‘0’ are transmitted as

2 Eb
S 1(t )  Cos 2f1t for symbol 1
Tb
S 2 (t )  0 for symbol 0

2. FSK[Frequency Shift Keying]:

In a binary FSK system symbol ‘1’ and ‘0’ are transmitted as

2 Eb
S 1(t )  Cos 2f1t for symbol 1
Tb
2 Eb
S 2 (t )  Cos 2f 2 t for symbol 0
Tb
3. PSK[Phase Shift Keying]:
In a binary PSK system the pair of signals S1(t) and S2(t) are used to represent binary
symbol ‘1’ and ‘0’ respectively.
2 Eb
S1 (t )  Cos 2f c t --------- for Symbol ‘1’
Tb

2 Eb 2 Eb
S 2 (t )  Cos(2f c t   )   Cos 2f c t ------- for Symbol ‘0’
Tb Tb

Hierarchy of digital modulation technique

Digital Modulation Technique

Coherent Non - Coherent

Binary M - ary Hybrid Binary M - ary

(m) = 2 (m) = 2
* ASK M-ary ASK M-ary APK * ASK M-ary ASK
* FSK M-ary FSK M-ary QAM * FSK M-ary FSK
* PSK M-ary PSK * DPSK M-ary DPSK
(QPSK)

Coherent Binary PSK:

Non Return to
Zero Level Product
Encoder Modulator
Binary Binary PSK Signal
Data Sequence

2
1 (t )  Cos 2f c t
Tb

Fig(a) Block diagram of BPSK transmitter

 x(t) Tb
x1 Decision Choose 1 if x1>0
 dt Device
0
Choose 0 if x1<0

Correlator
1 (t ) Threshold λ = 0

Fig (b) Coherent binary PSK receiver

In a Coherent binary PSK system the pair of signals S1(t) and S2(t) are used to represent
binary symbol ‘1’ and ‘0’ respectively.
2 Eb
S1 (t )  Cos 2f c t --------- for Symbol ‘1’
Tb

2 Eb 2 Eb
S 2 (t )  Cos(2f c t   )   Cos 2f c t ------- for Symbol ‘0’
Tb Tb

Eb 0  Eb1
Where Eb= Average energy transmitted per bit Eb 
2
In the case of PSK, there is only one basic function of Unit energy which is given by
2
1 (t )  Cos 2f c t 0  t  Tb
Tb
Therefore the transmitted signals are given by

S1 (t )  Eb 1 (t ) 0  t  Tb for Symbol 1

S 2 (t )   Eb 1 (t ) 0  t  Tb for Symbol 0

A Coherent BPSK is characterized by having a signal space that is one

dimensional (N=1) with two message points (M=2)

Tb

S11   S1 (t ) 1 (t ) dt   Eb
0

Tb

S 21   S 2 (t ) 1 (t ) dt   Eb
0
The message point corresponding to S1(t) is located at S11   Eb and S2(t) is located at
S 21   Eb .
To generate a binary PSK signal we have to represent the input binary sequence in polar
form with symbol ‘1’ and ‘0’ represented by constant amplitude levels of

 Eb &  Eb respectively. This signal transmission encoding is performed by a NRZ

level encoder. The resulting binary wave [in polar form] and a sinusoidal carrier 1 (t )

nc
[whose frequency f c  ] are applied to a product modulator. The desired BPSK wave
Tb
is obtained at the modulator output.
To detect the original binary sequence of 1’s and 0’s we apply the noisy PSK signal x(t)
to a Correlator, which is also supplied with a locally generated coherent reference signal
1 (t ) as shown in fig (b). The correlator output x 1 is compared with a threshold of zero
volt.
If x1 > 0, the receiver decides in favour of symbol 1.
If x1 < 0, the receiver decides in favour of symbol 0.

Probability of Error Calculation ‘Or’

Bit Error rate Calculation [BER Calculation] :-
In BPSK system the basic function is given by
2
1 (t )  Cos 2f c t 0  t  Tb
Tb
The signals S1(t) and S2(t) are given by

S1 (t )  Eb 1 (t ) 0  t  Tb for Symbol 1

S 2 (t )   Eb 1 (t ) 0  t  Tb for Symbol 0

The signal space representation is as shown in fig (N=1 & M=2)

Region R2 Region R1
 - Eb
Eb
0
Message Point 2 Message Point 1
S2(t) Decision S1(t)
Boundary
Fig. Signal Space Representation of BPSK
The observation vector x1 is related to the received signal x(t) by
T
x1   x(t )1 (t ) dt
0

If the observation element falls in the region R1, a decision will be made in favour
of symbol ‘1’. If it falls in region R2 a decision will be made in favour of symbol ‘0’.
The error is of two types
1) Pe(0/1) i.e. transmitted as ‘1’ but received as ‘0’ and
2) Pe(1/0) i.e. transmitted as ‘0’ but received as ‘1’.
Error of 1st kind is given by


1  ( x1   ) 2 
Pe (1 / 0)    2 2  dx1
2 2 0
exp Assuming Gaussian Distribution

Where μ = mean value =  Eb for the transmission of symbol ‘0’

N0
 2 = Variance = for additive white Gaussiance noise.
2
Threshold Value λ = 0. [Indicates lower limit in integration]

Therefore the above equation becomes

1
  ( x1  Eb ) 2 
N 0 0
Pe 0  Pe (1 / 0)  exp   dx1
 N0 

x1  Eb
Put Z 
N0


 exp(Z )  dz
1
Pe 0  Pe (1 / 0)  2

 ( Eb / N 0 )
 1 Eb
Pe (1 / 0)  erfc
2 N0

1 Eb
Similarly Pe (0 / 1)  erfc
2 N0

The total probability of error Pe  Pe (1 / 0) Pe (0)  Pe (0 / 1) Pe (1) assuming

probability of 1’s and 0’s are equal.
1
Pe  [ Pe (1 / 0)  Pe (0 / 1) ]
2
1 Eb
Pe  erfc
2 N0

Coherent Binary FSK

In a binary FSK system symbol ‘1’ and ‘0’ are transmitted as

2 Eb
S 1(t )  Cos 2f1t for symbol 1
Tb
2 Eb
S 2 (t )  Cos 2f 2 t for symbol 0
Tb

nc  i
Frequency f i  for some fixed integer nc and i=1, 2
Tb
The basic functions are given by
1
R c (k) 
N and

2
2 (t )  Cos 2f 2 t for 0  t  Tb and Zero Otherwise
Tb

Therefore FSK is characterized by two dimensional signal space with two message points
i.e. N=2 and m=2.
The two message points are defined by the signal vector

 E  0 
S1   b  and S2   
0   Eb 

Generation and Detection:-

 fig a

fig b

fig: FSK transmitter and receiver

A binary FSK Transmitter is as shown in fig. (a). The incoming binary

data sequence is applied to on-off level encoder. The output of encoder is Eb volts for

symbol 1 and 0 volts for symbol ‘0’. When we have symbol 1 the upper channel is
switched on with oscillator frequency f1, for symbol ‘0’, because of inverter the lower
channel is switched on with oscillator frequency f2. These two frequencies are combined
using an adder circuit and then transmitted. The transmitted signal is nothing but
required BFSK signal.
The detector consists of two correlators. The incoming noisy BFSK signal x(t) is
common to both correlator. The Coherent reference signal 1 (t ) and 2 (t ) are supplied to
upper and lower correlators respectively.
 The correlator outputs are then subtracted one from the other and resulting a
random vector ‘l’ (l=x1 - x2). The output ‘l’ is compared with threshold of zero volts.
If l > 0, the receiver decides in favour of symbol 1.
l < 0, the receiver decides in favour of symbol 0.

Probability of Error Calculation:-

In binary FSK system the basic functions are given by

2
1 (t )  Cos 2f1t 0  t  Tb
Tb

2
2 (t )  Cos 2f 2 t 0  t  Tb
Tb
The transmitted signals S1(t) and S2(t) are given by
S 1(t )  Eb 1 (t ) for symbol 1

S 2 (t )  E b  2 (t ) for symbol 0

Therefore Binary FSK system has 2 dimensional signal space with two messages S1(t)
and S2(t), [N=2 , m=2] they are represented as shown in fig.
Fig. Signal Space diagram of Coherent binary FSK system.

The two message points are defined by the signal vector

 E  0 
S1   b  and S2   
0   Eb 
The observation vector x1 and x2 ( output of upper and lower correlator) are
related to input signal x(t) as

Tb

x1   x(t )1 (t )dt and

0

Tb

x2   x(t )2 (t )dt

0

N0
Assuming zero mean additive white Gaussian noise with input PSD . with
2

N0
variance .
2
The new observation vector ‘l’ is the difference of two random variables x1 & x2.
l = x1 – x2

When symbol ‘1’ was transmitted x1 and x2 has mean value of 0 and Eb

respectively.
Therefore the conditional mean of random variable ‘l’ for symbol 1 was
transmitted is
l  x  x 
E   E 1   E 2 
1  1 1
 Eb  0
 Eb

l 
Similarly for ‘0’ transmission E     Eb
0
The total variance of random variable ‘l’ is given by
Var [l ]  Var [ x1 ]  Var [ x2 ]
 N0
The probability of error is given by

1
  (l  Eb ) 2 
Pe (1 / 0)  Pe 0 
2N 0 0 exp  2 N 0  dl
 

l  Eb
Put Z
2N 0

1
Pe 0   exp( z
2
)dz
 Eb
2 N0

1  Eb 
 erfc  
2  2 N 0 

1  Eb 
Similarly Pe1  erfc  
2  2 N 0 
1
The total probability of error = Pe  [ Pe (1 / 0)  Pe (0 / 1) ]
2
Assuming 1’s & 0’s with equal probabilities
1
Pe= [ Pe 0  Pe1 ]
2

1  Eb 
Pe  erfc  
2  2 N 0 

BINARY ASK SYSTEM:-

Binary ON-OFF Product Binary ASK

Level Modulator
Data Sequence Signal
Encoder
 2
1 (t )  Cos 2f e t
Tb

Fig (a) BASK transmitter

Tb
x(t) X  dt
0
Decision
Device
If x > λ choose symbol 1

If x < λ choose symbol 0

1 (t ) Threshold λ
Fig (b) Coherent binary ASK demodulator
In Coherent binary ASK system the basic function is given by

2
1 (t )  Cos 2f e t 0  t  Tb
Tb

The transmitted signals S1(t) and S2(t) are given by

S1 (t )  E b 1 (t ) for Symbol 1

S 2 (t )  0 for Symbol 0
The BASK system has one dimensional signal space with two messages (N=1, M=2)

Region E2 Region E1

Message
Point 2 Eb

1 (t )
Eb
0 Message
2
Point 1
Fig. (c) Signal Space representation of BASK signal

In transmitter the binary data sequence is given to an on-off encoder. Which

gives an output Eb volts for symbol 1 and 0 volt for symbol 0. The resulting binary
wave [in unipolar form] and sinusoidal carrier 1 (t ) are applied to a product modulator.
The desired BASK wave is obtained at the modulator output.
In demodulator, the received noisy BASK signal x(t) is apply to correlator with
coherent reference signal 1 (t ) as shown in fig. (b). The correlator output x is compared
with threshold λ.
If x > λ the receiver decides in favour of symbol 1.
If x < λ the receiver decides in favour of symbol 0.

BER Calculation:
In binary ASK system the basic function is given by

2
1 (t )  Cos 2f c t 0  t  Tb
Tb

The transmitted signals are given by

S1 (t )  E b  1(t ) for Symbol 1

S 2 (t )  0 for Symbol 0

A 2Tb
0
Therefore the average transmitted energy per bit Eb 0  Eb1 2
2  A Tb
Eb  
2 2 4
The probability of error is given by


1  ( x  0) 2 
Pe 0 
N 0   N o dx
exp
Eb
2

Where ‘x’ is the observed random vector. μ = 0, because the average value for symbol ‘0’
transmission is zero (0).

N0 N
2 assuming additive white Gaussian noise with into PSD 0
2 2
 x
Let Z 
N0


1
Pe 0   exp( z
2
)dz
 Eb
2 N0

1  Eb 
 erfc  
2  2 N 0 

1  Eb 
similarly Pe1  erfc  
2  2 N 0 
1
The total probability of error = [ Pe 0  Pe1 ]
2

1  Eb 
Pe  erfc  
2  2 N 0 

Incoherent detection:

Fig(a). : Envelope detector for OOK BASK

 Incoherent detection as used in analog communication does not require carrier for
reconstruction. The simplest form of incoherent detector is the envelope detector as
shown in figure(a). The output of envelope detector is the baseband signal. Once the
baseband signal is recovered, its samples are taken at regular intervals and compared with
threshold.
If Z(t) is greater than threshold (  ) a decision will be made in favour of symbol ‘1’
If Z(t) the sampled value is less than threshold (  ) a decision will be made in favour of
symbol ‘0’.
Non- Coherenent FSK Demodulation:-

Fig(b) : Incoherent detection of FSK

Fig(b) shows the block diagram of incoherent type FSK demodulator. The detector
consists of two band pass filters one tuned to each of the two frequencies used to
communicate ‘0’s and ‘1’s., The output of filter is envelope detected and then baseband
detected using an integrate and dump operation. The detector is simply evaluating which
of two possible sinusoids is stronger at the receiver. If we take the difference of the
outputs of the two envelope detectors the result is bipolar baseband.
The resulting envelope detector outputs are sampled at t = kTb and their values are
compared with the threshold and a decision will be made infavour of symbol 1 or 0.

Differential Phase Shift Keying:- [DPSK]

Fig. (a) DPSK Transmitter

Fig. (b) DPSK Receiver

A DPSK system may be viewed as the non coherent version of the PSK. It eliminates
the need for coherent reference signal at the receiver by combining two basic operations
at the transmitter
(1) Differential encoding of the input binary wave and
(2) Phase shift keying
Hence the name differential phase shift keying [DPSK]. To send symbol ‘0’ we
phase advance the current signal waveform by 1800 and to send symbol 1 we leave the
phase of the current signal waveform unchanged.
The differential encoding process at the transmitter input starts with an arbitrary
first but, securing as reference and thereafter the differentially encoded sequence{d k} is
generated by using the logical equation.

d k
 d b
k 1 k
 d b
k 1 k
 Where bk is the input binary digit at time kT b and dk-1 is the previous value of the
differentially encoded digit. Table illustrate the logical operation involved in the
generation of DPSK signal.

Input Binary Sequence {bK} 1 0 0 1 0 0 1 1

Differentially Encoded 1 1 0 1 1 0 1 1 1
sequence {dK}
Transmitted Phase 0 0 Π 0 0 Π 0 0 0
Received Sequence 1 0 0 1 0 0 1 1
(Demodulated Sequence)

A DPSK demodulator is as shown in fig(b). The received signal is first passed

through a BPF centered at carrier frequency fc to limit noise power. The filter output and
its delay version are applied to correlator the resulting output of correlator is proportional
to the cosine of the difference between the carrier phase angles in the two correlator
inputs. The correlator output is finally compared with threshold of ‘0’ volts .
If correlator output is +ve -- A decision is made in favour of symbol ‘1’
If correlator output is -ve --- A decision is made in favour of symbol ‘0’

COHERENT QUADRIPHASE – SHIFT KEYING

Fig. (a) QPSK Transmitter

Fig. (b) QPSK Receiver

In case of QPSK the carrier is given by

2E
si (t )  Cos[2f c t  (2i  1) / 4] 0t T i  1 to 4
T
2E 2E
si (t )  Cos[(2i  1) / 4] cos( 2f c t )  sin[(2i  1) / 4] sin( 2f c t ) 0  t  T i  1 to 4
T T
Fig. (c) QPSK Waveform

In QPSK system the information carried by the transmitted signal is contained in

the phase. The transmitted signals are given by

2E  
S (t ) 
1
T
cos 2

f c
t 
4
 for input di bit 10

2E  3 
S 2
(t ) 
T
cos 2

f c
t
4 
 for input dibit 00

2E  5 
S 3
(t ) 
T
cos 2

f c
t
4 
 for input dibit 01

2E  7 
S (t ) 
t
T
cos 2

f c
t
4 
 for input dibit 11

Where the carrier frequency f  nC for some fixed integer nc

C 7
E = the transmitted signal energy per symbol.
T = Symbol duration.

The basic functions  1 (t ) and  2 (t ) are given by

 (t ) 
2

cos 2 f t  0  t T
1
T b
c

 (t ) 
2

sin 2 f t  0  t T
2
T b
c

There are four message points and the associated signal vectors are defined by
  
 E cos  2i  1 4  
 
Si   i  1,2,3,4
   
 E sin  2i  1  
  4 

The table shows the elements of signal vectors, namely Si1 & Si2

Table:-

Input dibit Phase of Coordinates of message

QPSK points
Si1 Si2
signal(radians)

10   E  E
4 2 2
00 3  E  E
4 2 2
01 5  E  E
4 2 2
11 7  E  E
4 2 2
Therefore a QPSK signal is characterized by having a two dimensional signal
constellation(i.e.N=2)and four message points(i.e. M=4) as illustrated in fig(d)

.Fig (d) Signal-space diagram of coherent QPSK system.

Generation:-
Fig(a) shows a block diagram of a typical QPSK transmitter, the incoming binary
data sequence is first transformed into polar form by a NRZ level encoder. Thus the

symbols 1& 0 are represented by  E b

and  E b
respectively. This binary wave is

next divided by means of a demultiplexer [Serial to parallel conversion] into two separate
binary waves consisting of the odd and even numbered input bits. These two binary
waves are denoted by ao(t) and ae(t)
The two binary waves ao(t) and ae(t) are used to modulate a pair of quadrature

carriers or orthonormal basis functions  1 (t ) &  2 (t ) which are given by

2
 1
(t ) 
T
cos 2 f c
t

&
2
 2
(t ) 
T
sin 2 f c
t
 The result is a pair of binary PSK signals, which may be detected independently due to

the orthogonality of  (t ) & 

1 2
(t ) .
Finally the two binary PSK signals are added to produce the desired QPSK signal.
Detection:-
The QPSK receiver consists of a pair of correlators with a common input and supplied

with a locally generated pair of coherent reference signals  (t ) & 

1 2
(t ) as shown in

fig(b).The correlator outputs x1 and x2 produced in response to the received signal x(t) are
each compared with a threshold value of zero.
The in-phase channel output :
If x1 > 0 a decision is made in favour of symbol 1
x1 < 0 a decision is made in favour of symbol 0
Similarly quadrature channel output:
If x2 >0 a decision is made in favour of symbol 1 and
x2 <0 a decision is made in favour of symbol 0
Finally these two binary sequences at the in phase and quadrature channel outputs are
combined in a multiplexer (Parallel to Serial) to reproduce the original binary sequence.
Probability of error:-
A QPSK system is in fact equivalent to two coherent binary PSK systems
working in parallel and using carriers that are in-phase and quadrature.
The in-phase channel output x1 and the Q-channel output x2 may be viewed as the
individual outputs of the two coherent binary PSK systems. Thus the two binary PSK
systems may be characterized as follows.

- The signal energy per bit E

2

- The noise spectral density is N 0

2
The average probability of bit error in each channel of the coherent QPSK system is
  E 
P 
1 1
2
erfc 
 N 
2 E  E 2
 0

1  E 
 erfc  
2  2 N 0 

The bit errors in the I-channel and Q-channel of the QPSK system are statistically
independent . The I-channel makes a decision on one of the two bits constituting a
symbol (di bit) of the QPSK signal and the Q-channel takes care of the other bit.
Therefore, the average probability of a direct decision resulting from the
combined action of the two channels working together is
pc= probability of correct reception
p1= probability of error

P C
 1  P1  2

2
 1  E 
 1  erfc  
 2  2 No  
 E  1  E 
1  erfc    erfc 
2

 2 No  4  2 No 
The average probability of symbol error for coherent QPSK is given by

P e
 1 P C

 E  1  E 
 erfc    erfc 
2

 2 No  4  2 No 
E
In the region where 2 N   1 We may ignore the second term and so the
o

approximate formula for the average probability of symbol error for coherent QPSK
system is

E
P e
 erfc
2 No

Minimum shift keying:-

 In a continuous phase frequency shift keying [CPFSK] system the transmitted
signal is given by
 2
 E b cos 2
 T b
 f 1
t   (0)   for symbol 1
S (t )  
 2 Eb

 T b
cos 2  f 2
t   (0)   for symbol 0

Where Eb is the transmitted signal energy per bit and T b is bit duration the CPSK signal
S(t) is expressed in the conventional form of an angle modulated signal as

S (t ) 
2 Eb
cos 2 f t   (0) 
T b
c

The phase θ(t) is a continuous function of time which is given by

h t
 (t )   (0)  0 t  T
T b
b

The transmitted frequencies f 1

& f 2
are given by
h
f  f 
1 c
2T b

h
f  f 
2
2T c
b

f  1 / 2( f  f )
c 1 2

h T ( f  f )
b 1 2

Where f  the carrier frequency &

c

h  the deviation ratio

The variation of phase θ(t) with time t follows a path consisting of sequence of straight
lines, the slope of which represent frequency change Fig(a) shows the possible paths
starting from time t=0. This plot is known as “Phase tree”.
When h= ½ the frequency deviation equals half the bit rate. This is the minimum
frequency difference (deviation) that allows the two FSK signals representing symbol ‘1’
& ‘0’. For this reason a CPFSK signal with a deviation ratio of one- half is commonly
referred to as “minimum shift keying”[MSK].
Deviation ratio h is measured with respect to the bit rate 1/Tb
at t =Tb
h for Symbol 1
 (Tb )   (0)  
 h for Symbol 0

fig(b) phase tree

Fig(c): Phase Trellis, for sequence 1101000

In terms of In phase and Quadrature Component

2 Eb 2 Eb
s (t )  Cos [ (t )] Cos ( 2f c t )  Sin [ (t )] Sin ( 2f c t )
Tb Tb
with the deviation ratio h=1/2


 (t )   (0)  t 0  t  Tb
2Tb

+ Sign corresponds to symbol 1

- Sign corresponds to symbol 0
In phase components
For the interval of  Tb  t  Tb
consists of half cosine pulse
2 Eb
s1 (t )  Cos [  (t ) ]
Tb
2 Eb   
 Cos [  (0) ] Cos t 
Tb  2Tb 
2 Eb   
 Cos t   Tb  t  Tb
Tb  2Tb 

+ Sign corresponds to θ(0) =0

- Sign corresponds to θ(0) = п
Quadrature components
For the interval of 0  t  2 Tb
consists of half sine pulse

2 Eb
sQ (t )  Sin [  (t ) ]
Tb
2 Eb   
 Sin [  (Tb ) ] Cos t 
Tb  2Tb 
2 Eb   
 sin  t  0  t  2Tb
Tb  2Tb 

+ Sign corresponds to θ(Tb) =п/2

- Sign corresponds to θ(Tb) = -п/2
since the phase states θ(0) and θ(Tb) can each assume one of the two possible values,
any one of the four possibilities can arise
Basic functions are given by

2   
1 (t )  Cos  t  Cos (2f c t )  Tb  t  Tb
Tb  2Tb 

2   
2 (t )  Sin  t  Sin (2f ct ) 0  t  2Tb
Tb  2Tb 

we may express the MSK signal in the form

s(t )  s11 (t )  s22 (t ) 0  t  Tb

The coefficients are given by

Tb

s1   s (t ) 1 (t ) dt
Tb

 Eb Cos  (0)  Tb  t  Tb

and
2Tb

s2   s(t ) 2 (t ) dt
0b

  Eb Sin  (Tb ) 0  t  2Tb

The signal space diagram for MSK system is as shown in fig

Fig: signal space diagram for MSK system
Signal Space Characterization of MSK
Fig: sequence and waveforms for MSK signal
(a) input binary sequence (b) scaled time function s11 (t )

(c) scaled time function s 2 2 (t ) (d)obtained by adding (b) and (c)

In the case of AWGN channel, the received signal is given by
x(t) = s(t) + w(t)
where s(t) is the transmitted MSK signal and w(t) is the sample function of a white
Gaussian noise.
The projection of the received signal x(t) onto the reference signal 1 (t ) is

Tb

x1   x(t )  (t ) dt
Tb
1

 s1  w1  Tb  t  Tb
similarly the projection of the received signal x(t) onto the reference signal  2 (t ) is
2Tb

x2   x (t ) 2 (t ) dt
0

 s2  w2 0  t  2Tb

 Eb 
Pe  erfc  

 N 0 
 ^ 
If x2>0, the receiver chooses the estimate  (Tb )   . If, on the other hand, x2<0, it
2
^ 
chooses the estimate  (Tb )  .
2

To reconstruct the original binary sequence, we interleave the above two sets of
phase decisions,
^ ^ 
1 If we have the estimates  (0)  0 and  (Tb )   , or alternatively if we have the
2
^ ^ 
estimates  (0)   and  (Tb )  , the receiver makes a final decision in favor of
2
symbol 0.
^ ^ 
2 If we have the estimates  (0)   and  (Tb )   , or alternatively if we have
2
^ ^ 
the estimates  (0)  0 and  (Tb )  , the receiver makes a final decision in favor of
2
symbol 1.

Generation and detection of MSK signal:-

Fig. (b) MSK Receiver

Fig (a) shows the block diagram of typical MSK transmitter. and (b)receiver

n
Two input sinusoidal waves one of frequency f  C
for some fixed integer nc and
c
4T b

1
the other of frequency are first applied to a modulator. This produces two phase
4T b

coherent sinusoidal waves at frequencies f1 and f2 which are related to the carrier
frequency fc and the bit rate Rb by

h h
f  f  or f 
2 Rb
1 c
2T b
C

h h 1
f  f  or f 
2 Rb
for h 
2 c
2T b
C 2

These two sinusoidal waves are separated from each other by two narrow band filters one
centered at f1 and the other at f2. The resulting filter outputs are next linearly combined to

produce the pair of basis functions  (t )

1
and  (t ) . Finally
2
 (t )
1
and  (t ) are
2
multiplied with two binary waves a 1(t) and a2(t) both of which have a bit rate equal

1
to . These two binary waves are extracted from the incoming binary sequence.
2T b

Fig (b) shows the block diagram of a typical MSK receiver. The received signal
x(t) is correlated with locally generated replicas of the coherent reference signals

 (t )
1
and  (t ) . The integration in the Q – channel is delayed by Tb seconds with
2

respect to the I - channel.

The resulting in-phase and quadrature channel correlator outputs x 1 and x2 are
each compared with a threshold of zero. To estimate the phase θ(0) and θ(T b). Finally
these phase decisions are interleaved so as to reconstruct the original input binary
sequence with a minimum average probability of symbol error in an AGWN channel.
PROBLEM 1.
Binary data has to be transmitted over a telephone link that has a usable bandwidth of
3000Hz and a maximum achievable signal-to-noise power ratio of 6 dB at its output..
a. Determine the maximum signaling rate and probability of error if a coherent
ASK scheme is used for transmitting binary data through this channel.
b. If the data is maintained at 300 bits/sec, calculate the error probability.
Solution:
a) If we assume that an ASK signal requires a bandwidth of 3r b Hz, then the
maximum signaling rate permissible is given by
Bandwidth =3 rb=3000 Hz
rb=1000 bits/sec.

Average Signal Power  A2 4

Noise Power   2   2   3000 
Average Signal Power A2
4
Noise Power 12,000
A2
 48,000

Hence, A2 4rb  12 and
 
Pe  Q 12  Q 3.464   0.0003

b) If the bit rate is reduced to 300 bits/sec, then

 A2
 40
4rb
Pe  Q  
40  Q 6.326  1010

PROBLEM 2
Binary data is transmitted over an RF band pass channel with a usable bandwidth of
10 MHz at a rate of (4.8) (10 6) bits/sec using an ASK signaling method. The carrier
amplitude at the receiver antenna is 1 mv and the noise power spectral density at the
receiver input is 10-15 watt/Hz. Find the error probability of a coherent and non coherent
receiver..

Solution:
a) The bit error probability for the coherent demodulator is
 A 2T 
Pe  Q  b 
; A  1 mv, Tb  10 6 / 4.8
 4 
 
 / 2 10 watt / Hz
15


Pe  Q 26  2(10 7 ). 

b) for non coherent ASK pe is given by

 Pe  
1
2
  
exp  A 2Tb /(16 ) ,

pe = 0.0008

PROBLEM 3.
Binary data is transmitted at a rate of 106 bits/sec over a microwave link having a
bandwidth of 3 MHz. Assume that the noise power spectral density at the receiver input
is  / 2  10 10 watt / Hz. Find the average carrier power required at the receiver input for
coherent PSK and DPSK signaling schemes to maintain Pe ≤10-4.

Solution:

The probability of error for the PSK scheme is

 Pe  PSK  
 Q 2 S avTb /   10 4 ,
thus
2 S avTb /   3.75

( S av )  (3.75) 2 (10 10 ) (10 6 )  1.48dBm

For the DPSK scheme

  
 Pe  DPSK  1 exp  A 2Tb / 2 10 4 ,
2
Hence,
S av Tb /   8.517
 S av  DPSK  2.3.3 dBm

This example illustrates that the DPSK signaling scheme requires about 1 dB more power
than the coherent PSK scheme when the error probability is of the order of
10-4.
Probability of Error

Definition: Defines average probability of error that can occur in a Communication

system
Error Functions
(1) Error function erf(u):

u
2
erf (u )  
 0
exp( z 2 )dz -------- ( A6.1)

(2) Complementary error function erfc(u):


2
erfc(u )  
 u
exp( z 2 )dz -------- ( A6.2)

Properties of Error function

1. erf(-u) = -erf(u) - Symmetry.

2. erf(u) approaches unity as u tends towards infinity.

2
 exp( z )dz  1
2
-------- ( A6.3)
 0

3. For a Random variable X, with mean mx and variance σx2, the probability
of X is defined by

 a 
P(mX  a  X  m X  a)  erf   -------- ( A6.4)
 2 
 X 
Note: Relation: erfc(u) = 1 – erf(u)
Tables are used to find these values.

Approximate Relation: ( only for large values of u )

exp(u 2 )
erfc(u )  -------- ( A6.5)

Q – Function:
An alternate form of error function. It basically defines the area under the Standardized
Gaussian tail. For a standardized Gaussian random variable X of zero mean and unit
variance, the Q-function is defined by

1  x2 
Q (v ) 
2 v exp  2  dx -------- ( A6.6)

Relations between Q-function and erfc function:

1  v 
(i) Q (v )  erfc  ------- ( A6.7a)
2  2

(ii) erfc u   2Q( 2u ) ------ ( A6.7b)

Probability of Error Calculation for Binary PCM Systems
Consider a binary communication system in which the two symbols binary1 and binary0
are represented by the signals s1(t) and s2(t) respectively. Let E1 and E2 represent the
energies of the signals s1(t) and s2(t) respectively.
Tb Tb
E1   s (t ) dt2
1 and E2   s22 (t ) dt ----- ( A6.8)
0 0

The Probability of error for Communication Systems can be defined as

1  Eb ( 1   ) 
Pe  erfc 
 --------- ( A6.9)
2  2 N 0 
Where Eb is the average energy per bit defined by

E1  E 2
Eb  --------- ( A6.10)
2
and ρ is the correlation coefficient
Tb
1

Eb  s (t ) s (t ) dt
0
1 2 ------ ( A6.11)

and (No/2) represent the noise power spectral density in W/Hz.

Case (1): Uni-polar signaling:

In this scheme the signals are represented as

S1 (t )   a 0  t  Tb for Symbol 1

S 2 (t )  0 0  t  Tb for Symbol 0
Signal energies are E1 = a2 Tb and E2 = 0
Average energy per bit, Eb = a2 Tb/2.
Correlation coefficient = 0.
Probability of error,

1  Eb ( 1   ) 
Pe  erfc 
2  2 N 0 

1  a 2T 
Pe  erfc b 
2  4N0 
 
Case (2): Polar signaling:

In this scheme the signals are represented as

S1 (t )   a 0  t  Tb for Symbol 1

S 2 (t )  a 0  t  Tb for Symbol 0
Signal energies are E1 = a2 Tb and E2 = a2 Tb
Average energy per bit, Eb = a2 Tb
Correlation coefficient = -1.
Probability of error,

1  Eb (1   ) 
Pe  erfc 
2  2 N 0 

1  a 2T 
Pe  erfc b 
2  N0 
 
Case (3): Manchester signaling:
In this scheme the signals are represented as
S1 (t )   a / 2 0  t  Tb / 2 for Symbol 1
a/2 Tb / 2  t  Tb

S 2 (t )   a / 2 0  t  Tb / 2 for Symbol 0
a/2 Tb / 2  t  Tb
Signal energies are E1 = a2 Tb/4 and E2 = a2 Tb/4
Average energy per bit, Eb = a2 Tb/4
Correlation coefficient = -1.
Probability of error,

1  E (1   ) 
Pe  erfc b 
2  2 N 0 
Reduces to

1  a 2T 
Pe  erfc b 
2  4N0 
 
Example:

A binary PCM system using NRZ signaling operates just above the error threshold
with an average probability of error equal to 10-6. If the signaling rate is doubled,
find the new value of the average probability of error.

Solution:

For probability of error equal to 10-6.

Eb/N0 = 3.3 (from table)

The probability of error is

1  Eb 
Pe  erfc 

2  N0 
If the signaling rate is doubled then Eb is reduced by a factor of 2 and correspondingly
Eb/N0 also reduces by 2. Hence the new probability of error will become .

Pe  10 3
 Chapter 6

DETECTION AND ESTIMATION:

Fundamental issues in digital communications are

1. Detection and
2. Estimation

Detection theory: It deals with the design and evaluation of decision – making
processor that observes the received signal and guesses which particular symbol was
transmitted according to some set of rules.

Estimation Theory: It deals with the design and evaluation of a processor that uses
information in the received signal to extract estimates of physical parameters or
waveforms of interest.
The results of detection and estimation are always subject to errors

Model of digital communication system

Consider a source that emits one symbol every T seconds, with the symbols belonging to
an alphabet of M symbols which we denote as m1, m2, . . . . . . mM.
We assume that all M symbols of the alphabet are equally likely. Then
p i  p ( mi emitted )
1
 for all i
M
The output of the message source is presented to a vector transmitter producing vector of
real number
 S i1 
S 
 i 2 
 . 
Si    i  1, 2 , ..... , M
 . 
 . 
 

Where the dimension N ≤ M.

 S iN
 


The modulator then constructs a distinct signal si(t) of duration T seconds. The signal
si(t) is necessarily of finite energy.

The Channel is assumed to have two characteristics:

 Channel is linear, with a bandwidth that is large enough to accommodate
the transmission of the modulator output si(t) without distortion.
 The transmitted signal si(t) is perturbed by an additive, zero-mean,
stationary, white, Gaussian noise process.
such a channel is referred as AWGN ( additive white Gaussian noise ) channel

GRAM – SCHMIDT ORTHOGONALIZATION PROCEDURE:

In case of Gram-Schmidt Orthogonalization procedure, any set of ‘M’ energy signals

{Si(t)} can be represented by a linear combination of ‘N’ orthonormal basis functions
where N≤M. That is we may represent the given set of real valued energy signals S1(t),
S2(t). . . . . . . SM(t) each of duration T seconds in the form

S1 (t )  S111 (t )  S12 2 (t ) . . . . . . . .  S1N  N (t )

S 2 (t )  S 211 (t )  S 222 (t ) . . . . . . . .  S 2 N N (t )
.
.
.
.
S M (t )  S M 11 (t )  S M 22 (t ) . . . . . . . .  S MNN (t )

N
0  t  T
S i (t )   S ij  j (t ) 
j 1 i 1,2,3 ...... M  (6.1)

Where the Co-efficient of expansion are defined by

 T
i 1,2,3. . ... M
S ij (t )   S i (t ) j (t )dt 
0  j 1,2,3...... N  (6.2)

The basic functions 1 (t ),  2 (t ) ......... N (t ) are orthonormal by which

T
1 if i  j
0 i j
 (t ) (t ) dt  
0 if i  j  (6.3)
The co-efficient Sij may be viewed as the jth element of the N – dimensional
Vector Si
 S i1 
S 
 i2 
' 
 
Therefore S i  '  i = 1,2,3 . . . . . . M
' 
 
' 
S 
 iN 

Let S1  31 (t )  42 (t ) S 2   1 (t )  22 (t )

Vector 3    1
S1    S2   
 4  2

Geometric interpretation of signal:

Using N orthonormal basis functions we can represent M signals as
N
S i (t )   S ij  j (t ) 0 t  T i  1, 2 , ..... , M  (6.4)
j 1

Coefficients are given by

T
S ij   S i (t ) j (t ) dt i  1 , 2 , .. ... , M
0

j 1, 2 , ..... , N  (6.5)

Given the set of coefficients {sij}, j= 1, 2, ….N operating as input we may use the scheme
as shown in fig(a) to generate the signal si(t) i = 1 to M. It consists of a bank of N
multipliers, with each multiplier supplied with its own basic function, followed by a
summer.

fig(a)

conversely given a set of signals si(t) i = 1 to M operating as input we may use the
scheme shown in fig (b) to calculate the set of coefficients {sij}, j= 1, 2, ….N
 fig(b)

 S i1 
S 
 i 2 

The vector si is called signal vector

 . 
Si    i  1, 2 , ..... , M
 . 
 . 
 
 S iN
 


We may visualize signal vectors as a set of M points in an N dimensional Euclidean

space, which is also called signal space
The squared-length of any vector si is given by inner product or the dot product of si
N
 ( S i , Si )   S ij2
2
Si
j 1

Where sij are the elements of si

Two vectors are orthogonal if their inner product is zero

The energy of the signal is given by

T
Ei   S i2 (t ) dt
0

substituting the value si(t) from equation 6.1

T N N
Ei   [ S ij j (t )] [ S ikk (t )] dt
0 j 1 k 1

interchanging the order of summation and integration

N N T
Ei    S ij S ik   j (t )k (t ) dt
j 1 k 1 0
since  j (t ) forms an orthonormal set, the above equation reduce to

N
Ei   S ij2
j 1

this shows that the energy of the signal si(t) is equal to the squared-length of the signal
vector si
The Euclidean distance between the points represented by the signal vectors si and sk is
N
  ( S ij  S kj ) 2
2
Si  S k
j 1
T
  [ S i (t )  S k (t )]2 dt
0

Response of bank of correlators to noisy input

Received Signal X(t) is given by

X (t )  S i (t )  W (t ) 0t T
i  1,2,3. . . . . . ., M  (6.6)

where W(t) is AWGN with Zero Mean and PSD N0/2

Output of each correlator is a random variable defined by
T
X j   X (t ) j (t )dt
o

 S ij  W j j 1,2,........N  (6.7)

The first Component Sij is deterministic quantity contributed by the transmitted signal
Si(t), it is defined by
T
S ij  S
0
i (t ) j (t ) dt  (6.8)

The second Component Wj is a random variable due to the presence of the noise at the
input, it is defined by

T
W j   W (t ) j (t )dt  (6.9)
o
let X ' (t ) is a new random variable defined as
N
X ' (t )  X (t )  X j 1
j  j (t )  (6.10)

substituting the values of X(t) from 6.6 and Xj from 6.7 we get
N
X ' (t )  S i (t )  W (t )   (S
j 1
ij  W j ) j (t )

N
 W (t )  W 
j 1
j j (t )

 W ' (t )

which depends only on noise W(t) at the front end of the receiver and not at all on the
transmitted signal si(t). Thus we may express the received random process as
N
X (t )  X j 1
j  j (t )  X ' (t )
N
 X j 1
j  j (t )  W ' (t )

Now we may characterize the set of correlator output, {Xj}, j = 1 to N , since the received
random process X(t) is Gaussian , we deduce that each Xj is a Gaussian random variable.
Hence, each Xj is characterized completely by its mean and variance.

Mean and variance:

The noise process W(t) has zero mean, hence the random variable Wj extracted from W(t)
also has zero mean. Thus the mean value of he jth correlator output depends only on Sij as

m x j  E[ X j ] from eqn 6.7

 E[ S i j  W j ]
 S ij  E[W j ] but E[W j ]  0
 S ij

variance of Xj is given by
 2 x j  Var [ X j ]
 E[( X j  m x j ) 2 ] substituti ng m x j  S ij
 E[( X j  S ij ) 2 ] from equton 6.7
 E[W ] j
2
 substituting the value of Wj from eqn 6.9
T T

 2 x j  E   W (t )  j (t ) dt  W (u )  j (u ) du 
0 0 
T T 
 E     j (t )  j (u )W (t )W (u ) dt du 
0 0 

T T
 2x j   
0 0
j (t )  j (u ) E [ W (t )W (u ) ] dt du

T T
  
0 0
j (t )  j (u ) R w (t , u ) dt du  (6.11)

where

Rw (t , u )= E [ W (t )W (u ) ] autocorrelation function of the noise process W(t).Science the

noise is stationary, with psd N0/2 ,Rw(t,u) depends only on the time difference (t-u) and
expressed as

N0
Rw (t , u )   (t  u )  (6.12)
2
substituting this value in the equation 6.11 we get
T T
N0
 2x j 
2  
0 0
j (t )  j (u ) (t  u ) dt du

T
N0
 
2
j (t ) dt
2 0

Science the  j (t ) have unit energy, the above equation reduce to

N0
 2x j  for all j
2
This shows that all the correlator outputs {Xj}, j = 1 to N have a variance equal to the psd
No/2 of the additive noise process W(t).
Science the  j (t ) forms an orthogonal set, then the Xj are mutually uncorrelated, as
shown by
 Cov[ X j X k ]  E[( X j  mx j )( X k  mx k )]
 E[( X j  S i j )( X k  S ik )]
 E[W jWk ]
T T

 E   W (t ) j (t )dt  W (u )k (u )du 
0 0 
T T
  
0 0
j (t )k (u ) Rw (t , u )dt du

T T
N0

2  
0 0
j (t )k (u ) (t  u )dt du

T
N0

2 
0
 j (t )k (u )dt

0 jk

Since the Xj are Gaussian random variables, from the above equation it is implied that
they are also statistically independent.

Detection of known signals in noise

Assume that in each time slot of duration T seconds, one of the M possible signals S1(t),
S2(t). . . . . . . SM(t) is transmitted with equal probability of 1/M. Then for an AWGN
channel a possible realization of sample function x(t), of the received random process
X(t) is given by

x (t )  Si (t )  w(t ) 0t T
i  1,2,3,.............., M

where w(t) is sample function of the white Gaussian noise process W(t), with zero mean
and PSD N0/2. The receiver has to observe the signal x(t) and make a best estimate of
the transmitted signal si(t) or equivalently symbol mi
The transmitted signal si(t), i= 1to M , is applied to a bank
of correlators, with a common input and supplied with an appropriate set of N
orthonormal basic functions, the resulting correlator outputs define the signal vector Si.
knowing Si is as good as knowing the transmitted signal Si(t) itself, and vice versa. We
may represents si(t) by a point in a Euclidean space of dimensions N ≤ M. . Such a point
is referred as transmitted signal point or message point. The collection of M message
points in the N Euclidean space is called a signal constellation.
When the received signal x(t) is applied to the bank o N correlators , the output of the
correlator define a new vector x called observation vector. this vector x differs from the
signal vector si by a random noise vector w
x  Si  w i  1,2,3,........., M
The vectors x and w are sampled values of the random vectors X and W respectively. the
noise vector w represents that portion of the noise w(t) which will interfere with the
detected process.
Based on the observation vector x, we represent the received signal s(t)by a point in the
same Euclidean space, we refer this point as received signal point. The relation between
them is as shown in the fig

Fig: Illustrating the effect of noise perturbation on location of the received signal point

In the detection problem , the observation vector x is given, we have to perform a

mapping from x to an estimate of the transmitted symbol, in away that would minimize
the average probability of symbol error in the decision. The maximum likelihood
detector provides solution to this problem.

Optimum transmitter & receiver

 Probability of error depends on signal to noise ratio

 As the SNR increases the probability of error decreases
 An optimum transmitter and receiver is one which maximize the SNR and
minimize the probability of error.
Correlative receiver

Observation
Vector x

For an AWGN channel and for the case when the transmitted signals are equally likely,
the optimum receiver consists of two subsystems
1) .Receiver consists of a bank of M product-integrator or correlators
Φ1(t) ,Φ2(t) …….ΦM(t) orthonormal function
The bank of correlator operate on the received signal x(t) to produce observation vector x
2). Implemented in the form of maximum likelihood detector that
operates on observation vector x to produce an estimate of the transmitted symbol
mi i = 1 to M, in a way that would minimize the average probability of symbol error.
The N elements of the observation vector x are first multiplied by the corresponding N
elements of each of the M signal vectors s1, s2… sM , and the resulting products are
successively summed in accumulator to form the corresponding set of
Inner products {(x, sk)} k= 1, 2 ..M. The inner products are corrected for the fact that the
transmitted signal energies may be unequal. Finally,
the largest in the resulting set of numbers is selected and a corresponding decision on the
transmitted message made.
The optimum receiver is commonly referred as a correlation receiver

MATCHED FILTER
Science each of t he orthonormal basic functions are Φ1(t) ,Φ2(t) …….ΦM(t) is assumed to
be zero outside the interval 0  t  T . we can design a linear filter with impulse
response hj(t), with the received signal x(t) the fitter output is given by the convolution
integral

y j (t )   x( ) h

j (t   ) d

Suppose the impulse response of the system is

h j (t )   j (T  t )

Then the filter output is


y j (t )   x ( )

j (T  t   ) d
sampling this output at time t = T, we get

y j (T )   x( )

j ( ) d

Φj(t) is zero outside the interval 0  t  T ,we get

T
y j (T )   x ( ) j ( )d
0

yj(t) = xj
where xj is the j th correlator output produced by the received signal x(t).

A filter whose impulse response is time-reversed and delayed version of the input signal
 j (t ) is said to be matched to  j (t ) . correspondingly , the optimum receiver based on
this is referred as the matched filter receiver.

For a matched filter operating in real time to be physically realizable, it must be causal.
For causal system

h j (t )  0 t 0

causality condition is satisfied provided that the signal  j (t ) is zero outside the interval
0t T

Maximization of output SNR in matched filter

Let
x(t) = input signal to the matched filter
h(t) = impulse response of the matched filter
w(t) =white noise with power spectral density No/2
 (t ) = known signal
Input to the matched filter is given by
x (t )   (t )  w(t ) 0t T
science the filter is linear , the resulting output y(t) is given by
y (t )   0 (t )  n(t )

where 0 (t ) and n(t) are produced by the signal and noise components of the input x(t).

The signal to noise ratio at the output of the matched filter at t = T is

2
 0 (T )
( SNR ) 0   (6.13)
E[ n 2 (t )]

aim is to find the condition which maximize the SNR

let
0 (t )  0 ( f )
h(t )  H ( f )
are the Fourier transform pairs, hence the output signal  0 (t ) is given by

0 (t ) 

 H ( f ) ( f ) exp( j 2ft )df

output at t = T is
 2


2
 0 (T )  H ( f ) ( f ) exp( j 2fT ) df  (6.14)

 For the receiver input noise with psd No/2 the receiver output noise psd is given by
N0 2
S N( f )  H ( f )  (6.15)
2
and the noise power is given by

E[n 2 (t )]  S

N ( f ) df


N0
 H( f )
2
 df  (6.16)
2 

substituting the values of eqns 6.14 & 6.15

2
in 6.13 we get


 H ( f ) ( f ) exp( j 2fT )df


( SNR ) 0  
 (6.17)
N0

2
H ( f ) df
2 

using Schwarz’s inequality

 2  

X  
2 2
1 ( f ) X 2 ( f ) df  X1( f ) df X2( f ) df  (6.18)
  

Eqn 6.16 is equal when X1(f) = kX2*(f)

let X1(f) = H(f)
& X2(f) =  ( f ) exp( j 2fT )
under equality condition

H(f) = K  * ( f ) exp( j 2fT )  (6.19)

Thus substituting in 6.16 we get the value

 2  

 H ( f )( f ) exp( j 2fT )df  H( f )  ( f )

2 2
 df df
  

substituting in eqn 6,17 and simplifying


2
 ( f )
2
( SNR ) 0  df
N0 

Using Rayleigh’s energy theorem

 

  (t )  ( f )
2 2
dt  df  E , energy of the signal
 

2E
( SNR ) 0, max     (6.20)
N0
 Under maximum SNR condition, the transfer function is given by ( k=1), eqn 6.19

H opt ( f )   * ( f ) exp( j 2fT )

The impulse

response in time domain is given by
hopt (t )    (  f ) exp[ j 2f (T

 t )] exp( j 2ft ) df

  (T  t )

Thus the impulse response is folded and shifted version of the input signal  (t )

MATCHED FILTER

Φ(t) = input signal

h(t) = impulse response
W(t) =white noise

The impulse response of the matched filter is time-reversed and delayed version of the
input signal
h (t )   (T  t )
For causal system
h j (t )  0 t 0

Matched filter properties

PROPERTY 1
The spectrum of the output signal of a matched filter with the matched signal as
input is, except for a time delay factor, proportional to the energy spectral density of
the input signal.
let  0 ( f ) denotes the Fourier transform of the filter output  0 (t ) , hence

 0 ( f )  H opt ( f )  ( f ) sustitutin g from 6.19

  * ( f )  ( f ) exp( j 2fT )
2
  ( f ) exp(2 jfT )  (6.21)

PROPERTY 2

The output signal of a Matched Filter is proportional to a shifted version of the

autocorrelation function of the input signal to which the filter is matched.
The autocorrelation function and energy spectral density of a signal forms the Fourier
transform pair, thus taking inverse Fourier transform for eqn 6.21
0 (t )  R (t  T )

At time t = T

0 (T )  R (0)  E

where E is energy of the signal

PROPERTY 3

The output Signal to Noise Ratio of a Matched filter depends only on the ratio of the
signal energy to the power spectral density of the white noise at the filter input.

SNR at the output of matched filter is eqn 6.13

2
 0 (T )
( SNR ) 0   (6.22)
E[ n 2 (t )]
 output of matched filter is

0 (t )   H ( f ) ( f ) exp( j 2ft )df


signal power at t = T

 2

 H ( f ) ( f ) exp( j 2fT ) df
2
0 (T ) 


noise psd at the output of receiver is

N0 2
S N( f )  H( f )
2
noise power is
 
N0
S  H( f )
2
E[ n 2 (t )]  N ( f ) df  df

2 

substituting the values in 6.22 we get

 2

 H ( f ) ( f ) exp( j 2fT )df


( SNR ) 0  
 (6.23)
N0

2
H( f ) df
2 

using Schwarz’s inequality

 2  

X  
2 2
1 ( f ) X 2 ( f ) df  X1( f ) df X2( f ) df  (6.24)
  

Eqn 6.24 is equal when X1(f) = kX2*(f)

let X1(f) = H(f)
& X2(f) =  ( f ) exp( j 2fT )
under equality condition

H(f) = K  * ( f ) exp( j 2fT )  (6.25)

Thus substituting in 6.24 we get the value

 2  

 H ( f )( f ) exp( j 2fT )df  H( f )  ( f )

2 2
 df df
  
 substituting in eqn 6,23 and simplifying

2
 ( f )
2
( SNR ) 0  df
N0 

Using Rayleigh’s energy theorem

 

  (t )  ( f )
2 2
dt  df  E , energy of the signal
 

2E
( SNR ) 0, max   (6.26)
N0

PROPERTY 4

The Matched Filtering operation may be separated into two matching conditions;
namely spectral phase matching that produces the desired output peak at time T,
and the spectral amplitude matching that gives this peak value its optimum signal to
noise density ratio.

In polar form the spectrum of the signal  (t ) being matched may be expressed as

 ( f )   ( f ) exp  j ( f ) 

where  ( f ) is magnitude spectrum and  ( f ) is phase spectrum of the signal.

The filter is said to be spectral phase matched to the signal  (t ) if the transfer function of
the filter is defined by

H ( f )  H ( f ) exp   j ( f )  j 2fT 

The output of such a filter is


 0' (t )   H ( f ) ( f ) exp( j 2ft )df


 

H ( f )  ( f ) exp[ j 2f (t  T )]df

The product H ( f ) ( f ) is real and non negative.

The spectral phase matching ensures that all the spectral components of the output add
constructively at t = T , there by causing the output to attain its maximum value.


0' (t )  0' (T )  

 ( f ) H ( f ) df

For spectral amplitude matching

H ( f )  ( f )

Problem-1:
Consider the four signals s1(t), s2(t), s3(t) and s4(t) as shown in the fig-P1.1 .
Use Gram-Schmidt Orthogonalization Procedure to find the orthonormal
basis for this set of signals. Also express the signals in terms of the basis functions.

Fig-P1.1: Signals for the problem -1.

Solution:

Given set is not linearly independent because s4(t) = s1(t) + s3(t)

Step-1: Energy of the signal s1(t)

T
T
E1   s12 (t ) dt 
0
3
s1 (t ) 3
First basis function 1 (t )   for 0  t  T
E1 T 3
Step-2: Coefficient s21
T
s 21  0
s 2 (t ) 1 (t ) dt  T
3
T
2T
Energy of s2(t) E 2   s 22 (t ) dt 
0
3
s2 (t )  s21 1 (t ) 3
Second Basis function 2 (t )   for T  t  2T
E2  s 2 T 3 3
21

T
Step-3: Coefficient s31: s31  0
s3 (t ) 1 (t ) dt  0

T
Coefficient s32 s32  0
s3 (t ) 2 (t ) dt  T
3
Intermediate function
g3(t) = s3(t) - s31Φ1(t) - s32 Φ2(t)

g3(t) = 1 for 2T/3 < t < T/3

Third Basis function

g 3 (t ) 3
3 (t )   for 2T  t T
T T 3

2
g (t ) dt
3
0
The corresponding orthonormal functions are shown in the figure-P1.2.
 Fig-P1.2: Orthonormal functions for the Problem-1

Representation of the signals

S1 (t )  T 1 (t )
3

S 2 (t )  T 1 (t )  T 3 2 (t )
3

S 3 (t )  T 2 (t )  T 3 3 (t )
3

S 4 (t )  T 1 (t )  T 3 2 (t )  T 3 3 (t )
3
PROBLEM-2:

Consider the THREE signals s 1(t), s2(t) and s3(t) as shown in the fig P2.1. Use
Gram-Schmidt Orthogonalization Procedure to find the orthonormal basis for this
set of signals. Also express the signals in terms of the basis functions.

Fig-P2.1: Signals for the problem -2.

Solution: The basis functions are shown in fig-P2.2.

Fig-P2.2: Orthonormal functions for the Problem-2

Correspondingly the representation of the signals are:

S1 (t )  2 1 (t )

S 2 (t )   4 1 (t )  4 2 (t )

S3 (t )  3 1 (t )  3 2 (t )  3 3 (t )
PROBLEM-3:

Consider the signal s(t) in fig-P3.1

a) Determine the impulse
response of a filter matched
to this signal and sketch it as
a function of time.
b) Plot the matched filter output
as a function of time.
c) What is Peak value of the
output?

Fig P3.1
Solution:

The impulse response of the matched filter is time-reversed and delayed version of the
input signal, h(t) = s(T-t) and the output of the filter, y(t) = x(t) * h(t).

Given s(t) = +1 for 0 < t < 0.5

-1 for 0.5 < t < 1.
(a) With T = 1, the impulse
response h(t) is

h(t) = -1 for 0 < t < 0.5

+1 for 0.5 < t < 1.
 Fig. P3.2

(b) The output of the filter y(t) is obtained by convolving the input s(t) and the
impulse response h(t). The corresponding output is shown in the fig. P3.3.

(c) The peak value of the output is 1.0 unit.

Fig. P3.3

Assignment Problem:

Specify a matched filter for the signal S1(t) shown in Fig.-P4.1 Sketch the output of
the filter matched to the signal S1(t) is applied to the filter input.

Fig P4.1

126
e-Notes by M.N.Suma, BMS, Bangalore

CHAPTER-4
Line Codes

In base band transmission best way is to map digits or symbols into pulse
waveform. This waveform is generally termed as Line codes.
RZ: Return to Zero [ pulse for half the duration of Tb ]
NRZ Return to Zero[ pulse for full duration of Tb ]

1 0 1 0 1 1 1 0 0

Unipolar
NRZ

Polar NRZ

NRZ-inverted
(differential
encoding)

Bipolar
encoding

Manchester
encoding

Differential
Manchester
encoding

Unipolar (NRZ)

Unipolar NRZ

127
Unipolar NRZ
“1” maps to +A pulse “0” maps to no pulse

 Poor timing
 Low-frequency content
 Simple
 Long strings of 1s and 0s ,synchronization problem

Polar - (NRZ)

Polar NRZ
“1” maps to +A pulse “0” to –A pulse
 Better Average Power
 simple to implement
 Long strings of 1s and 0s ,synchronization problem
 Poor timing

Bipolar Code

NRZ-
Bipolar
V
1 0 1 0 0 1 1 0 1
0

• Three signal levels: {-A, 0, +A}

• “1” maps to +A or –A in alternation
• “0” maps to no pulse
• Long string of 0’s causes receiver to loose synchronization

128
 • Suitable for telephone systems.
Manchester code

1 0 1 0 1 1 1 0 0

Manchester
Encoding

• “1” maps into A/2 first for Tb/2, and -A/2 for next Tb/2
• “0” maps into -A/2 first for Tb/2, and A/2 for Tb/2
• Every interval has transition in middle
– Timing recovery easy
• Simple to implement
• Suitable for satellite telemetry and optical communications

Differential encoding
 It starts with one initial bit .Assume 0 or 1.
 Signal transitions are used for encoding.

Example NRZ - S and NRZ – M

 NRZ –S : symbol 1 by no transition , Symbol 0 by transition.
 NRZ-M : symbol 0 by no transition , Symbol 1 by transition
 Suitable for Magnetic recording systems.

M-ary formats
Bandwidth can be properly utilized by employing M-ary formats. Here
grouping of bits is done to form symbols and each symbol is assigned some
level.
Example
Polar quaternary format employs four distinct symbols formed by dibits.
Gray and natural codes are employed

Parameters in choosing formats

1. Ruggedness
2. DC Component
3. Self Synchronization.
4. Error detection
5. Bandwidth utilization
6. Matched Power Spectrum

129
Power Spectra of Discrete PAM Signals:
The discrete PAM signals can be represented by random process

X(t)  K   A k V(t  KT)



Where Ak is discrete random variable, V(t) is basic pulse, T is symbol

duration. V(t) normalized so that V(0) = 1.
Coefficient Ak represents amplitude value and takes values for different
line codes as
Symbol 1 a
A  [ Symbol 0  0
k
Unipolar

Symbol 1 a
A  [ Symbol 0  a
k
Polar
Alternate Symbol 1 takes  a, a
Bipolar A
k
 [ Symbol 0  0

Manchester Symbol 1 a
A  [ Symbol 0  a
k

As Ak is discrete random variable, generated by random process X(t),

We can characterize random variable by its ensemble averaged auto correlation
function given by
RA(n) = E [Ak.Ak-n] ,
Ak, Ak-n = amplitudes of kth and (k-n)th symbol position

PSD & auto correlation function form Fourier Transform pair & hence auto
correlation function tells us something about bandwidth requirement in frequency
domain.

Hence PSD Sx(f) of discrete PAM signal X(t).is given by

1 2   j2π fnT
Sx (f)  V(f)  R A (n) e
T n 

130
Where V(f) is Fourier Transform of basic pulse V(t). V(f) & R A(n) depends on
different line codes.

Power Spectra of NRZ Unipolar Format

Consider unipolar form with symbol 1’s and 0’s with equal probability i.e.
P(Ak=0) = ½ and P(Ak=1) = ½

For n=0;
Probable values of Ak.Ak = 0 x 0 & a x a

=E [ Ak.Ak-0]
= E[Ak2] = 02 x P [ Ak=0] + a2 x P[Ak=1]
RA(0) = a2/2

If n ≠ 0
Ak.Ak-n will have four possibilities (adjacent bits)
0 x 0, 0 x a, a x 0, a x a with probabilities ¼ each.

E[Ak.Ak-n] = 0 x ¼ + 0 x ¼ + 0 x ¼ + a2 / 4
= a2 / 4
V(t) is rectangular pulse of unit amplitude, its Fourier Transform will be
sinc function.

V(f) = FT [ V(t)] = Tb Sinc (fTb) PSD is given by

1 2   j2π fnT
Sx (f)  V(f)  R A (n) e
T n 
substituting the values of V(f) and RA(n)

  j2π fnT
S (f)  1 T 2 Sinc2 (fT )  R (n) e b
X T  b b  n  A
b

 
 
   j2π fnT 
 2 (fT )  R (0)  b
Tb Sinc
 b 
  A  R A (n) e 
 n  
 

 n 0 


 
 

2 (fT )  a 2  a 2   j2π fnTb 

Tb Sinc e 
 b 
 2 4 
 n  
 

 n 0 


131
 2 2   j2π fn Tb
 a T Sinc2 (fT )  a T Sinc2 (fT )  e
4 b b 4 b b n

using Poisson’s formula

  j2π fn Tb 
e  1  δ(f - n )
n  T T
b n  b

2 2 
S (f)  a T Sinc2 (fT )  a T Sinc2 (fT ) 1  δ(f  n )
X 4 b b 4 b b T n  T
b b

 n
 δ(f  T ) is Dirac delta train which multiplies Sinc function which
n b
1 2
has nulls at  ,  . . . . . . . . . .
Tb Tb

 n
As a result, Sin 2 (fTb ).  δ(f  )  δ(f)
n  Tb

Where δ(f) is delta function at f = 0,

Therefore a 2T 2
S (f)  b Sin 2 (fT )  a δ(f )
X 4 b 4

Power Spectra of Bipolar Format

Here symbol 1 has levels ± a, and symbol 0 as 0. Totally three levels.

Let 1’s and 0’s occur with equal probability then

P(AK = a) = 1/4 For Symbol 1

P(AK = -a) = 1/4
P(AK = 0) = 1/2 For Symbol 0

For n=0

132
 E[AK2] = a x a P(AK = a) + (0 x 0) P[AK = 0] +
(-a x –a) P(AK = -a)
= a2/4 + 0 + a2/4 = a2/2

For n≠0, i.e. say n=1;

Four possible forms of AK.AK-1
00,01,10,11 i.e. dibits are
0 X 0, 0 X ±a, ±a X 0, ±a X ±a
with equal probabilities ¼.
E[AK.AK-1] = 0 x ¼ + 0 x ¼ + 0 x ¼ - a2 x ¼
= -a2/4

For n>1, 3 bits representation 000,001,010 . . . . . . 111. i.e. with each probability
of 1/8 which results in
E[AK.AK-n] = 0

a2 / 2 n=0
Therefore RA(n) = -a2 / 4 n = ±1
0 n>1

1 2   j2π fnT
Sx (f)  V(f)  R A (n) e
T n 

PSD is given by
1 T 2SinC 2 (fT )  R ( 1)e j2π fnTb   R (0)  R (1)e- j2π fnTb 
S x (f) 
T  b b   A  A  A 
b

b

Sx (f)  T SinC 2 (fT ) 
b  2

 a 2 a 2 j2π fnTb

4
(e e
 j2π fTb 
)



 a 2T 
Sx (f)   b SinC 2 (fT ) 1  Cos(2fTb ) 
 2 b 
 

 a 2T
 

S x (f)   b SinC 2 (fT ) 2Sin 2 ( fTb)
 2 b 
 

 
S x (f)  a 2 T SinC 2 (fT ) Sin 2 ( fTb)
b b


133
Spectrum of Line codes

• Unipolar most of signal power is centered around origin and there is

waste of power due to DC component that is present.
• Polar format most of signal power is centered around origin and they are
simple to implement.
• Bipolar format does not have DC component and does not demand more
bandwidth, but power requirement is double than other formats.
• Manchester format does not have DC component but provides proper
clocking.
Spectrum suited to the channel.
• The PSD of the transmitted signal should be compatible with the
channel frequency response
• Many channels cannot pass dc (zero frequency) owing to ac
coupling
• Low pass response limits the ability to carry high frequencies
Inter symbol Interference

Generally, digital data is represented by electrical pulse, communication channel

is always band limited. Such a channel disperses or spreads a pulse carrying
digitized samples passing through it. When the channel bandwidth is greater

134
than bandwidth of pulse, spreading of pulse is very less. But when channel
bandwidth is close to signal bandwidth, i.e. if we transmit digital data which
demands more bandwidth which exceeds channel bandwidth, spreading will
occur and cause signal pulses to overlap. This overlapping is called Inter
Symbol Interference. In short it is called ISI. Similar to interference caused by
other sources, ISI causes degradations of signal if left uncontrolled. This
problem of ISI exists strongly in Telephone channels like coaxial cables and
optical fibers.
In this chapter main objective is to study the effect of ISI, when digital data
is transmitted through band limited channel and solution to overcome the
degradation of waveform by properly shaping pulse.
1 0 11

Tb
Transmitted Waveform Pulse Dispersion

The effect of sequence of pulses transmitted through channel is shown in

fig. The
Spreading of pulse is greater than symbol duration, as a result adjacent pulses
interfere. i.e. pulses get completely smeared, tail of smeared pulse enter into
adjacent symbol intervals making it difficult to decide actual transmitted pulse.
First let us have look at different formats of transmitting digital data.In
base band transmission best way is to map digits or symbols into pulse
waveform. This waveform is generally termed as Line codes.

BASEBAND TRANSMISSION:

135
PAM signal transmitted is given by


x(t)   a V(t  KTb ) -------------------------- (1)
K  K

V(t) is basic pulse, normalized so that V(0) = 1,

x(t) represents realization of random process X(t) and a k is sample value of
random variable ak which depends on type of line codes.
The receiving filter output

y(t)  μ a
K  
k P(t  KTb ) ---------------(2)

The output pulse μ P(t) is obtained because input signal ak .V(t) is passed
through series of systems with transfer functions HT(f), HC(f), HR(f)

Therefore μ P(f) = V(f). HT(f).HC(f).HR(f) --------- (3)

P(f) p(t) and V(f) v(t)

The receiving filter output y(t) is sampled at ti = iTb. where ‘i’ takes intervals
i = ±1, ±2 . . . . .


y (iTb )   a
K  
k P (iT b  KTb )


y (iTb )   ai P (0)   a
K  
k P (iT b  KTb ) ---------------------(4)

136
 K=i K≠i

In equation(4) first term μai represents the output due to ith transmitted bit.
Second term represents residual effect of all other transmitted bits that are
obtained while decoding i th bit. This unwanted residual effect indicates ISI. This
is due to the fact that when pulse of short duration T b is transmitted on band
limited channel, frequency components of the pulse are differentially attenuated
due to frequency response of channel causing dispersion of pulse over the
interval greater than Tb.
In absence of ISI desired output would have y (ti) = μai

Nyquist Pulse Shaping Criterion

In detection process received pulse stream is detected by sampling at

intervals ±KTb, then in detection process we will get desired output. This
demands sample of ith transmitted pulse in pulse stream at K th sampling interval
should be

P(iTb – KTb) = 1 K=i

0 K≠i ------------- (5)
If received pulse P(t) satisfy this condition in time domain, then
y(ti) = μai
Let us look at this condition by transform eqn(5) into frequency domain.
Consider sequence of samples {P(nTb)} where n=0,±1. . . . . . . by
sampling in time domain, we write in frequency domain


1
p ( f ) 
Tb
 p( f  n / T
n  
b ) ----------------(6)

Where pδ(f) is Fourier transform of an infinite period sequence of delta

functions of period Tb but pδ(f) can be obtained from its weighted sampled P(nT b)
in time domain

 
p ( f )    p(mT
  m  
b )  (t  mTb ) e  j 2ft dt  p (t ). (t )

137
 Where m = i-k, then i=k, m=0; so

p ( f )  

p (0)  (t ) e  j 2ft dt

Using property of delta function



i.e

  (t ) dt  1

Therefore p ( f )  p (0)  1

Pδ(f) = 1 ------------(7)
p(0) =1 ,i.e pulse is normalized (total area in frequency domain is unity)

Comparing (7) and (6)

1 
 p( f  n / Tb )  1
Tb n 


1
Or  p( f  n / T )  T
n  
b b 
Rb
----------- (8)

Where Rb = Bit Rate

Is desired condition for zero ISI and it is termed Nyquist’s first criterion for
distortion less base band transmission. It suggests the method for constructing
band limited function to overcome effect of ISI.

Ideal Solution
Ideal Nyquist filter that achieves best spectral efficiency and avoids ISI is
designed to have bandwidth as suggested

B0 = 1/2Tb (Nyquist bandwidth) = Rb/2

ISI is minimized by controlling P(t) in time domain or P(f) to be rectangular

function in frequency domain.

1  f 
P(f) = rect  
2B0  2B0 

 1 if f < B0

P(f) =  2B0
0 f > B0


138
Impulse response in time domain is given by
sin(2π B0 t)
P(t) 
2π B0 t
 sinc(2B t)
0

Disadvantage of Ideal solution

• P(f) to be flat from –Bo to +Bo and zero else where , abrupt transition is
physically not realizable.
• For large values of ‘ t’ , function P(t) decreases as resulting in slower
decay of sinc function due to discontinuity of P(f)
This causes timing error which results in ISI.

Practical solution
Raised Cosine Spectrum
• To design raised cosine filter which has transfer function consists of a flat
portion and a roll off portion which is of sinusoidal form
1
• Bandwidth 
B 2 is an adjustable value between B o and 2Bo.
Tb
0

 1 f  f1
 2Bo

P(f) = 
 1 f1 f  2Bof 1

 


 π f f1

 
 4Bo1cos  

   2Bo2f1  
 
  
 


 0 f  2Bo  f1

139
The frequency f1 and bandwidth Bo are related by
α 1  f 1

B 0 α is called the roll off factor

for α = 0, f1=Bo and BW=Bo is the minimum Nyquist bandwidth for the .
rectangular spectrum.
• For given Bo , roll off factor ‘α’ specifies the required excess bandwidth
• α =1,indicates required excess bandwidth is 100% as roll off
characteristics of P(f) cuts off gradually as compared with ideal low pass
filters. This function is practically realizable.
Impulse response P(t) is given by

cos(2π αBot)
P(t)  sinc(2Bot)
1  16 α Bo 2 t 2

140
P(t) has two factors

• sinc(2Bot) which represents ideal filter - ensures zero crossings

• 1
second factor that decreases as 2
- helps in reducing tail of sinc pulse
i.e. fast decay t

P(t) 

sinc 4 B0 t 
• For α =1, 1  16 B0 2 t 2

At t= Tb p(t)=0.5
2
Pulse width measured exactly equal to bit duration T b. Zero crossings occur at
t = ±3Tb, ±5Tb… In addition to usual crossings at t = ±Tb, ±2Tb… Which helps in
time synchronization at receiver at the expense of double the transmission
bandwidth

Transmission bandwidth required can be obtained from the relation

B = 2Bo- f1

Where B = Transmission bandwidth

Bo = 1Nyquist bandwidth
2Tb

f1
But α = 1-
B0
using
f1 = B0 (1- α)

B = 2 B0 – B0(1- α)

therefore B = B0(1+ α)

α =0; B=B0, minimum band width

α =1; B=2B0 , sufficient bandwidth

141
Roll-off factor
Smaller roll-off factor:
• Less bandwidth, but
• Larger tails are more sensitive to timing errors
Larger roll-off factor:
• Small tails are less sensitive to timing errors, but
• Larger bandwidth
Example1

A certain telephone line bandwidth is 3.5Khz .calculate data rate in bps

that can be transmitted if binary signaling with raised cosine pulses and
roll off factor α = 0.25 is employed.

Solution:
α = 0.25 ---- roll off
B = 3.5Khz ---transmission bandwidth

B = Bo(1+ α)
1 Rb
B0 =  Ans: Rb= 5600bps
2Tb 2
Example2

A source outputs data at the rate of 50,000 bits/sec. The transmitter uses
binary PAM with raised cosine pulse in shaping of optimum pulse width.
Determine the bandwidth of the transmitted waveform. Given
a.α = 0 b.α = 0.25 c. α = 0.5 d. α = 0.75 e. α = 1

Solution
B = B0(1+ α) B0=Rb/2
a. Bandwidth = 25,000(1 + 0) = 25 kHz
b. Bandwidth = 25,000(1 + 0.25) = 31.25 kHz
c. Bandwidth = 25,000(1 + 0.5) = 37.5 kHz
d. Bandwidth = 25,000(1 + 0.75) = 43.75 kHz
e. Bandwidth = 25,000(1 + 1) = 50 kHz

Example 3

142
A communication channel of bandwidth 75 KHz is required to transmit
binary data at a rate of 0.1Mb/s using raised cosine pulses. Determine the
roll off factor α.
Rb = 0.1Mbps
B=75Khz
α=?

B = Bo(1+ α)
B0 = Rb/2 Ans : α =0.5

Correlative coding :
So far we treated ISI as an undesirable phenomenon that produces a
degradation in system performance, but by adding ISI to the transmitted signal in
a controlled manner, it is possible to achieve a bit rate of 2Bo bits per second in a
channel of bandwidth Bo Hz. Such a scheme is correlative coding or partial-
response signaling scheme. One such example is Duo binary signaling.
Duo means transmission capacity of system is doubled.

Duo binary coding

143
Consider binary sequence {bk} with uncorrelated samples transmitted at the rate
of Rb bps. Polar format with bit duration T b sec is applied to duo binary
conversion filter. when this sequence is applied to a duobinary encoder, it is
converted into three level output, namely -2, 0 and +2.To produce this
transformation we use the scheme as shown in fig.The binary sequence {b k} is
first passed through a simple filter involving a single delay elements. For every
unit impulse applied to the input of this filter, we get two unit impulses spaced T b
seconds apart at the filter output. Digit C k at the output of the duobinary encoder
is the sum of the present binary digit bk and its previous value bk-1
Ck = bk + bk-1

The correlation between the pulse amplitude C k comes from bk and previous bk-1
digit, can be thought of as introducing ISI in controlled manner., i.e., the
interference in determining {b k} comes only from the preceding symbol {b k-1} The
symbol {bk} takes ±1 level thus Ck takes one of three possible values -2,0,+2 .
The duo binary code results in a three level output. in general, for M-ary
transmission, we get 2M-1 levels
Transfer function of Duo-binary Filter

The ideal delay element used produce delay of T b seconds for impulse will have
transfer function e -j 2π f Tb .
Overall transfer function of the filter H(f)

144
  j 2 π f Tb
H(f)  H c (f)  H c (f)e

  j2 π f T b 
H(f)  H c (f) 1 e 
 

 j πfT  j πfT b 
e b
e   j πfTb
 2 H c (f) e

2 
 

 j π f Tb
 2Hc (f)cos(π
As ideal channel transfer function
f Tb) e
 1
1 f 
Hc (f)   2 Tb
0 otherwise


Thus overall transfer function

  j π f Tb 1
2cos(π f Tb) e f 
 2 Tb
H(f)  

0 otherwise




H(f) which has a gradual roll off to the band edge, can also be implemented by
practical and realizable analog filtering Fig shows Magnitude and phase plot of
Transfer function

145
Advantage of obtaining this transfer function H(f) is that practical implementation
is easy
Impulse response
Impulse response h(t) is obtained by taking inverse Fourier transformation of
H(f)

 j 2π f t df
h(t)   H(f)e


1
2 Tb
 jπ f Tb
  2 cos( π f T b)e [ e j 2π f t ] df
 1
2 Tb

 πt   π t  T b  
sin  sin 
 
  Tb    Tb 
 πt   π t  T b  
   
T 
 b  Tb 
 πt   πt 
sin  
 sin  
  Tb   Tb 
 πt   π t  T b  
   
T  Tb
 b   146
  πt 
T 2sin  
b Tb
h(t ) 
πt  Tb  1

Impulse response has two sinc pulses displaced by T b sec. Hence overall
impulse response has two distinguishable values at sampling instants t = 0 and t
= Tb .

Overall Impulse response

147
Duo binary decoding

Encoding : During encoding the encoded bits are given by

Ck = bk + bk-1

Decoding:

At the receiver original sequence {b k} may be detected by subtracting the

previous decoded binary digit from the presently received digit C k This
demodulation technique (known as nonlinear decision feedback equalization) is
essentially an inverse of the operation of the digital filter at the transmitter

if
^
bk
is estimate of original sequence b then
k

^   ^
bk Ck bk 1
Disadvantage

If Ck and previous estimate

^ is received properly without error then we
b k -estimate
get correct decision and current 1 Otherwise once error made it tends to
propagate because of decision feed back. current {b k} depends on previous bk-1.

148
Example consider sequence 0010110

Precoding
In case of duo binary coding if error occurs in a single bit it reflects as multiple
errors because the present decision depends on previous decision also. To
make each decision independent we use a precoder at the receiver before
performing duo binary operation.
The precoding operation performed on the input binary sequence {b k}
converts it into another binary sequence {ak} given by

a k  b k  a k 1

a modulo 2 logical addition

Unlike the linear operation of duo binary operation, the precoding is a non
linear operation.

149
Fig. A precoded duo binary scheme.

{ak} is then applied to duobinary coder, which produce sequence {C k}

C k  a k  a k 1

If that symbol at precoder is in polar format Ck takes three levels,

 2v if
 b k  symbol 0
Ck  
0v
 if b k  symbol 1

The decision rule for detecting the original input binary sequence {b k} from {ck} is

 symbol 0 if Ck  1v
b ^k  
 symbol 1 if C k  1v

150
Example: with start bit as 0, reference bit 1

Example: with start bit as 0, reference bit 0

151
Example: with start bit as 1, reference bit 1

Example: with start bit as 1, reference bit 0

152
 Today the duo-binary techniques are widely applied throughout the world.
While all current applications in digital communications such as data
transmission, digital radio, and PCM cable transmission, and other new
possibilities are being explored.
This technique has been applied to fiber optics and to high density disk recording
which have given excellent results

Example

The binary data 001101001 are applied to the input of a duo binary system.

a)Construct the duo binary coder output and corresponding receiver output,
without a precoder.
b) Suppose that due to error during transmission, the level at the receiver input
produced by the second digit is reduced to zero. Construct the new receiver
output.
c) Repeat above two cases with use of precoder

without a precoder

errors errors

153
With a precoder (start bit 1)

With a precoder (start bit 0)

154
 The Transfer function H(f) of Duo binary signalling has non zero spectral
value at origin, hence not suitable for channel with Poor DC response. This
drawback is corrected by Modified Duobinary scheme.

Modified Duobinary scheme.

It is an extension of the duo-binary signaling. The modified duo binary

technique involves a correlation span of two binary digits. Two-bit delay causes
the ISI to spread over two symbols. This is achieved by subtracting input binary
digits spaced 2Tb secs apart.

Modified Duobinary scheme.

155
Transmitter

The precoded output sequence is given by

a modulo 2 logical addition a b  a
k k k 2

If ak = ± 1v , Ck takes one of three values 2,0,-2.

Output sequence of modified duo binary filter is given by Ck

Ck = ak – ak-2

Ck takes one of three values 2,0,-2

Ck = 0V, if bk is represented by symbol 0

Ck = +2V, if bk is represented by symbol 1

Receiver
At the receiver we may extract the original sequence {b k} using the decision rule
bk = symbol 0 if |Ck| >1v
symbol 1 if |Ck| ≤1v

The Transfer function of the filter is given by

156
 - j 4 π f Tb
H(f)  Hc(f) - Hc(f)e

 - j 4 π f Tb 
 Hc(f) 1- e 
 
  j 2 π f Tb - j 2 π f Tb 
- j 2 π f Tb  e  e 
 2 jHc(f) e
 2j 
 
- j 2 π f Tb
 2 jHc(f) sin (2π f Tb) e
 1
1 f 
Where H c (f) is  2 Tb
0 otherwise


 - j 2 π f Tb 1
2j sin ( 2 π f T b) e f 
H(f)  
 2 Tb
0 Otherwise


The Transfer function has zero value at origin, hence suitable for poor dc
channels

Impulse response

Impulse response h(t) is obtained by taking Inverse Fourier transformation of H(f)

 j 2π f t df
h(t)   H(f)e

157
 1
2 Tb
 j π f Tb
  2 jsin ( 2π f T b)e [ e j 2π f t ] df
 1
2 Tb
 πt   π t  2T b  
sin   sin  
 T b   Tb 
 
 πt   π t  2T b  
   
T  Tb
 b  
 πt    πt  
sin   sin  
 Tb   Tb 
 
 πt   π t  2T b  
   
T  Tb
 b  
 π t  
2Tb 2 sin 
  Tb 
πt  2T b has
Impulse response t  three distinguishable levels at the sampling instants.

158
To eliminate error propagation modified duo binary employs Precoding option
same as previous case.

Prior to duo binary encoder precoding is done using modulo-2 adder on signals
spaced 2Tb apart

a b  a
k k k 2

Example : Consider binary sequence {bk}= {01101101} applied to input of a

precoded modified duobinary filter. Determine receiver output and compare
with original {bk}.

159
Consider binary sequence {bk}={01101101}

Example

The binary data 011100101 are applied to the input of a modified duo binary
system.
a) Construct the modified duobinary coder output and corresponding receiver
output, without a precoder.
b) Suppose that due to error during transmission, the level at the receiver input
produced by the third digit is reduced to zero. Construct the new receiver output.
c) Repeat above two cases with use of precoder

Modified duobinary coder output and corresponding receiver output, without a

precoder

160
Modified duo binary coder output and corresponding receiver output, with a
precoder

161
Generalized form of correlative coding scheme

Fig. Generalized Correlative Coding

162
 The Duo binary and modified Duo binary scheme have correlation spans
of one binary digit and two binary digits respectively. This generalisation scheme
involves the use of a tapped delay line filter with tap weights f0, f1, f2,… fn-1. A
correlative samples Ck is obtained from a superposition of ‘N’ successive input
sample values bK
N 1

Ck = f
n 0
n b k -n

By choosing various combination values for f n, different correlative coding

schemes can be obtained from simple duo binary.

Base band Transmission of M-ary data

In base band M-ary PAM, output of the pulse generator may take on any one of
the M-possible amplitude levels with M>2 for each symbol
The blocks of n- message bits are represented by M-level waveforms with
M=2n.

Ex: M=4 has 4 levels. possible combination are 00, 10, 11, 01

T= 2Tb is termed symbol duration .

In general symbol duration T = Tb log 2M.

M-ary PAM system is able to transmit information at a rate of log2M faster than
binary PAM for given channel bandwidth.

Rb
R
log2 M

Rb = bit rate for binary system

R = symbol rate for binary system

M-ary PAM system requires more power which is increased by factor equal to

M2
log2 M for same average probability of symbol error.

M-ary Modulation is well suited for the transmission of digital data over channels
that offer a limited bandwidth and high SNR

Example

163
An analog signal is sampled, quantised and encoded into a binary PCM
wave. The number of representation levels used is 128. A synchronizing
pulse is added at the end of each code word representing a sample of the
analog signal. The resulting PCM wave is transmitted over a channel of
bandwidth 12kHz using binary PAM system with a raised cosine spectrum.
The roll off factor is unity.
a)Find the rate (in BPS ) at which information is transmitted through the
channel.
b) Find the rate at which the analog signal is sampled. What is the
maximum possible value for the highest frequency component of the
analog signal.

Solution
Given Channel with transmission BW B=12kHz.
Number of representation levels L = 128
Roll off α = 1

a) B = Bo(1+ α),
Hence Bo =6kHz.
Bo=Rb/2 therefore Rb = 12kbps.

b) For L=128, L = 2n , n = 7
symbol duration T = Tb log2M =nTb
sampling rate fs = Rb/n = 12/7 = 1.714kHz.
And maximum frequency component of analog signal is
From LP sampling theorem w = fs/2 = 857Hz.

Eye pattern

The quality of digital transmission systems are evaluated using the bit
error rate. Degradation of quality occurs in each process modulation,
transmission, and detection. The eye pattern is experimental method that
contains all the information concerning the degradation of quality. Therefore,
careful analysis of the eye pattern is important in analyzing the degradation
mechanism.

164
 • Eye patterns can be observed using an oscilloscope. The received wave
is applied to the vertical deflection plates of an oscilloscope and the
sawtooth wave at a rate equal to transmitted symbol rate is applied to the
horizontal deflection plates, resulting display is eye pattern as it resembles
human eye.
• The interior region of eye pattern is called eye opening

We get superposition of successive symbol intervals to produce eye pattern as

shown below.

• The width of the eye opening defines the time interval over which the
received wave can be sampled without error from ISI

• The optimum sampling time corresponds to the maximum eye opening

• The height of the eye opening at a specified sampling time is a measure

of the margin over channel noise.

165
The sensitivity of the system to timing error is determined by the rate of closure
of the eye as the sampling time is varied.

Any non linear transmission distortion would reveal itself in an asymmetric or

squinted eye. When the effected of ISI is excessive, traces from the upper
portion of the eye pattern cross traces from lower portion with the result that the
eye is completely closed.

Example of eye pattern:

Binary-PAM Perfect channel (no noise and no ISI)

Example of eye pattern: Binary-PAM with noise no ISI

Example 1

A binary wave using polar signaling is generated by representing symbol 1 by a

pulse of amplitude -1v; in both cases the pulse duration equals the bit duration.
The signal is applied to a low pass RC filter with transfer function
1
H(f) 
1  jf/f0

Construct eye pattern for the filter for

1. Alternating 1s and 0s
2. A long sequence of 1s followed by long sequence of zeros.

166
Example 2

The binary sequence 011010 is transmitted through channel having a raised

cosine characteristics with roll off factor unity. Assume the use of polar signaling,
format.
construct the Eye pattern

167
 0 1 1 0 1 0
Eye Pattern

Adaptive equalization for data transmission

This technique is another approach to minimize signal distortion in the

base band data transmission. This is Nyquist third method for controlling ISI.
Equalization is essential for high speed data transmission over voice
grade telephone channel which is essentially linear and band limited.
High speed data transmission involves two basic operations:
i) Discrete pulse amplitude modulation:
The amplitudes of successive pulses in a periodic pulse train are varied in
a discrete fashion in accordance with incoming data stream.
ii) Linear modulation:
Which offers band width conservation to transmit the encoded pulse train
over telephone channel.
At the receiving end of the systems, the received waves is demodulated
and then synchronously sampled and quantized. As a result of dispersion of the
pulse shape by the channel the number of detectable amplitude levels is limited
by ISI rather than by additive noise. If the channel is known , then it is possible to
make ISI arbitrarily small by designing suitable pair of transmitting and receiving
filters for pulse shaping.
In switched telephone networks we find that two factors contribute to pulse
distortion.
1. Differences in the transmission characteristics of individual links that may be
switched together.

168
2. Differences in number of links in a connection
Because of these two characteristics, telephone channel is random in nature.
To realize the full transmission capability of a telephone channel we need
adaptive equalization.

Adaptive equalization

• An equalizer is a filter that compensates for the dispersion effects of a

channel. Adaptive equalizer can adjust its coefficients continuously during
the transmission of data.

Pre channel equalization

 requires feed back channel
 causes burden on transmission.

Post channel equalization

Achieved prior to data transmission by training the filter with the guidance
of a training sequence transmitted through the channel so as to adjust the filter
parameters to optimum values.

Adaptive equalization – It consists of tapped delay line filter with set of delay
elements, set of adjustable multipliers connected to the delay line taps and a
summer for adding multiplier outputs.

169
The output of the Adaptive equalizer is given by
M1
y(nt)  c
i  0
i x(nT  iT)

Ci is weight of the ith tap Total number of taps are M .Tap spacing is equal to
symbol duration T of transmitted signal
In a conventional FIR filter the tap weights are constant and particular designed
response is obtained. In the adaptive equaliser the Ci's are variable and are
adjusted by an algorithm

Two modes of operation

1. Training mode 2 . Decision directed mode

Mechanism of adaptation

Training mode

A known sequence d(nT) is transmitted and synchronized version of it is

generated in the receiver applied to adaptive equalizer.

This training sequence has maximal length PN Sequence, because it has large
average power and large SNR, resulting response sequence (Impulse) is
observed by measuring the filter outputs at the sampling instants.

The difference between resulting response y(nT) and desired response d(nT)is
error signal which is used to estimate the direction in which the coefficients of
filter are to be optimized using algorithms

170
Methods of implementing adaptive equalizer

i) Analog
ii) Hard wired digital
iii) Programmable digital

Analog method

 Charge coupled devices [CCD’s] are used.

 CCD- FET’s are connected in series with drains capacitively coupled to
gates.

 The set of adjustable tap widths are stored in digital memory locations,
and the multiplications of the analog sample values by the digitized tap
weights done in analog manner.

 Suitable where symbol rate is too high for digital implementation.

Hard wired digital technique

• Equalizer input is first sampled and then quantized in to form that is

suitable for storage in shift registers.

• Set of adjustable lap weights are also stored in shift registers. Logic
circuits are used for required digital arithmetic operations.

• widely used technique of equalization

Programmable method

• Digital processor is used which provide more flexibility in adaptation by

programming.

• Advantage of this technique is same hardware may be timeshared to

perform a multiplicity of signal processing functions such as filtering,
modulation and demodulation in modem.

171
 CHAPTER-7
Spread – Spectrum Modulation
Introduction:
Initially developed for military applications during II world war, that was less
sensitive to intentional interference or jamming by third parties.
Spread spectrum technology has blossomed into one of the fundamental building
blocks in current and next-generation wireless systems
Problem of radio transmission

Narrow band can be wiped out due to interference

To disrupt the communication, the adversary needs to do two things,
(a) to detect that a transmission is taking place and
(b) to transmit a jamming signal which is designed to confuse the receiver.
.Solution
A spread spectrum system is therefore designed to make these tasks as difficult
as possible.
Firstly, the transmitted signal should be difficult to detect by an
adversary/jammer, i.e., the signal should have a low probability of intercept (LPI).
Secondly, the signal should be difficult to disturb with a jamming signal, i.e., the
transmitted signal should possess an anti-jamming (AJ) property
Remedy
spread the narrow band signal into a broad band to protect against
interference
In a digital communication system the primary resources are Bandwidth
and Power. The study of digital communication system deals with efficient
utilization of these two resources, but there are situations where it is necessary to
sacrifice their efficient utilization in order to meet certain other design objectives.
For example to provide a form of secure communication (i.e. the
transmitted signal is not easily detected or recognized by unwanted listeners) the
bandwidth of the transmitted signal is increased in excess of the minimum
bandwidth necessary to transmit it. This requirement is catered by a technique
known as “Spread Spectrum Modulation”.

172
 The primary advantage of a Spread – Spectrum communication system is
its ability to reject ‘Interference’ whether it be the unintentional or the intentional
interference.
The definition of Spread – Spectrum modulation may be stated in two
parts.
1. Spread Spectrum is a mean of transmission in which the data sequence
occupies a BW (Bandwidth) in excess of the minimum BW necessary to
transmit it.
2. The Spectrum Spreading is accomplished before transmission through
the use of a code that is independent of the data sequence. The Same
code is used in the receiver to despread the received signal so that the
original data sequence may be recovered.

s(t) wide band r(t) wide band b(t) +

Noise
b(t) ..... ..... Narrow
Wide
Band
Band
c(t) n(t) c(t)
Wide band (noise) Wide band

---- Transmitter---- ---- Channel------ --- Receiver--------

fig:1 spread spectrum technique.
b(t) = Data Sequence to be transmitted (Narrow Band)
c(t) = Wide Band code
s(t) = c(t) * b(t) – (wide Band)

173
Fig: Spectrum of signal before & after spreading
PSUEDO-NOISE SEQUENCE:
Generation of PN sequence:

Clock
Shift Shift Shift Output
Register1 Register2 Register3
S0 S1 S 2 S3 Sequence

Logic Circuit
Fig 2: Maximum-length sequence generator for n=3

A feedback shift register is said to be Linear when the feed back logic
consists of entirely mod-2-address ( Ex-or gates). In such a case, the zero state
is not permitted. The period of a PN sequence produced by a linear feedback
shift register with ‘n’ flip flops cannot exceed 2 n-1. When the period is exactly 2n-
1, the PN sequence is called a ‘maximum length sequence’ or ‘m-sequence’.
Example1: Consider the linear feed back shift register as shown in fig
2 involve three flip-flops. The input so is equal to the mod-2 sum of S1 and S3. If
the initial state of the shift register is 100. Then the succession of states will be
as follows.
100,110,011,011,101,010,001,100 . . . . . .

174
 The output sequence (output S3) is therefore. 00111010 . . . . .
Which repeats itself with period 23–1 = 7 (n=3)
Maximal length codes are commonly used PN codes
In binary shift register, the maximum length sequence is
N = 2m-1

chips, where m is the number of stages of flip-flops in the shift register.

At each clock pulse

• Contents of register shifts one bit right.
• Contents of required stages are modulo 2 added and fed back
to input.

Fig: 1
Initial stages
0 of
0 Shift 0registers 1000
0 1 0 0
Let initial
0 status
0 of1shift register
0 be 1000
1 0 0 1
1 1 0 0
0
1
1
0
1
1
0
1
• We can see for shift Register of length
m=4. .At each clock the change in state of flip-
0 1 0 1
flop is shown.
1 0 1 0
1
1
1
1
0
1
1
0
• Feed back function is modulo two of X3
and X4.
1 1 1 1
0
0
1
0
1
1
1
1
• After 15 clock pulses the sequence
175
repeats.
0 0 0 1
Output sequence is
1 0 0 0
000100110101111
Properties of PN Sequence
Randomness of PN sequence is tested by following properties
1. Balance property
2. Run length property
3. Autocorrelation property
1. Balance property
In each Period of the sequence , number of binary ones differ from binary zeros
by at most one digit .
Consider output of shift register 0 0 0 1 0 0 1 1 0 1 0 1 1 1 1
Seven zeros and eight ones -meets balance condition.
2. Run length property
Among the runs of ones and zeros in each period, it is desirable that about one
half the runs of each type are of length 1, one- fourth are of length 2 and one-
eighth are of length 3 and so-on.
Consider output of shift register
Number of runs =8

0 0 0 1 0 0 1 1 0 1 0 1 1 1 1
3 1 2 2 1 1 1 4

3. Auto correlation property

Auto correlation function of a maximal length sequence is periodic and binary
valued. Autocorrelation sequence of binary sequence in polar format is given by

1 N
R c (k)   c c
N n 1 n n -k

176
Where N is length or period of the sequence and
k is the lag of the autocorrelation
 1 if k l N

R c (k)   1
 k  l N
 N

Where l is any integer. 1

Rc (k) 
we can also state Autocorrelation function as N
{ No. of agreements – No. of disagreements in comparison of one full period }
Consider output of shift register for l=1

1
Rc (k)  (7  8)
15
1
R c (k) 
15

Yields PN autocorrelation as

Range of PN Sequence Lengths

Length 0f Shift Register, m PN Sequence Length,

7 127
8 255
9 511

177
 10 1023
11 2047
12 4095
13 8191
17 131071
19 524287
A Notion of Spread Spectrum:
An important attribute of Spread Spectrum modulation is that it can
provide protection against externally generated interfacing signals with finite
power. Protection against jamming (interfacing) waveforms is provided by
purposely making the information – bearing signal occupy a BW far in excess of
the minimum BW necessary to transmit it. This has the effect of making the
transmitted signal a noise like appearance so as to blend into the background.
Therefore Spread Spectrum is a method of ‘camouflaging’ the information –
bearing signal.
V
b(t) m(t). . r(t) z(t)
Tb Decisio
 dt
0
n
Device

c(t) n(t) c(t)

Threshold=0
<----Transmitter-----> --Channel -  - - - - Receiver
-------------

Let { bK} denotes a binary data sequence.

{ cK } denotes a PN sequence.
b(t) and c(t) denotes their NRZ polar representation respectively.
The desired modulation is achieved by applying the data signal b(t) and PN
signal c(t) to a product modulator or multiplier. If the message signal b(t) is
narrowband and the PN sequence signal c(t) is wide band, the product signal

178
m(t) is also wide band. The PN sequence performs the role of a ‘Spreading
Code”.
For base band transmission, the product signal m(t) represents the
transmitted signal. Therefore m(t) = c(t).b(t)
The received signal r(t) consists of the transmitted signal m(t) plus an
additive interference noise n(t), Hence
r(t) = m(t) + n(t)
= c(t).b(t) + n(t)

+1

0
-1
a) Data Signal b(t)
+1

0
-1

b)Spreading Code c(t)

+1

179
 0
-1

c)Product signal or base band transmitted signal m(t)

To recover the original message signal b(t), the received signal r(t) is applied to a
demodulator that consists of a multiplier followed by an integrator and a decision
device. The multiplier is supplied with a locally generated PN sequence that is
exact replica of that used in the transmitter. The multiplier output is given by
Z(t) = r(t).c(t)
= [b(t) * c(t) + n(t)] c(t)
= c2(t).b(t) + c(t).n(t)
The data signal b(t) is multiplied twice by the PN signal c(t), where as unwanted
signal n(t) is multiplied only once. But c2(t) = 1, hence the above equation
reduces to
Z(t) = b(t) + c(t).n(t)
Now the data component b(t) is narrowband, where as the spurious component
c(t)n(t) is wide band. Hence by applying the multiplier output to a base band (low
pass) filter most of the power in the spurious component c(t)n(t) is filtered out.
Thus the effect of the interference n(t) is thus significantly reduced at the receiver
output.
The integration is carried out for the bit interval 0 ≤ t ≤ T b to provide the sample
value V. Finally, a decision is made by the receiver.
If V > Threshold Value ‘0’, say binary symbol ‘1’
If V < Threshold Value ‘0’, say binary symbol ‘0’

Direct – Sequence Spread Spectrum with coherent binary Phase shift Keying:-
Binary data
Binary PSK
b(t) m(t) x(t)
Modulator
c(t)
PN Code Carrier
Generator

180
 a) Transmitter

Say 1
Tb Decision
y(t) Coherent v v
if v > 0 Detector  dt
0
Device

Received Say 0
Signal c(t) if v < 0
Local PN
generator

Local Carrier

b) Receiver

Fig: model of direct – sequence spread binary PSK system(alternative form)

To provide band pass transmission, the base band data sequence is

multiplied by a Carrier by means of shift keying. Normally binary phase shift
keying (PSK) is used because of its advantages.
The transmitter first converts the incoming binary data sequence {b k} into
an NRZ waveform b(t), which is followed by two stages of modulation.

181
 The first stage consists of a multiplier with data signal b(t) and the PN
signal c(t) as inputs. The output of multiplier is m(t) is a wideband signal. Thus a
narrow – band data sequence is transformed into a noise like wide band signal.
The second stage consists of a binary Phase Shift Keying (PSK)
modulator. Which converts base band signal m(t) into band pass signal x(t). The
transmitted signal x(t) is thus a direct – sequence spread binary PSK signal. The
phase modulation θ(t) of x(t) has one of the two values ‘0’ and ‘π’ (180 o)
depending upon the polarity of the message signal b(t) and PN signal c(t) at time
t.
Polarity of PN & Polarity of PN signal both +, + or - - Phase ‘0’
Polarity of PN & Polarity of PN signal both +, - or - + Phase ‘π’
Polarity of data sequence b(t)
+ -
Polarity of PN sequence C(t) + 0 π
- π 0
The receiver consists of two stages of demodulation.
In the first stage the received signal y(t) and a locally generated carrier are
applied to a coherent detector (a product modulator followed by a low pass filter),
Which converts band pass signal into base band signal.
The second stage of demodulation performs Spectrum despreading by
multiplying the output of low-pass filter by a locally generated replica of the PN
signal c(t), followed by integration over a bit interval T b and finally a decision
device is used to get binary sequence.
Signal Space Dimensionality and Processing Gain
 Fundamental issue in SS systems is how much protection spreading
can provide against interference.
 SS technique distribute low dimensional signal into large dimensional
signal space (hide the signal).
 Jammer has only one option; to jam the entire space with fixed total
power or to jam portion of signal space with large power.

Consider set of orthonormal basis functions;

182
  2
 cos(2π fct) kTc  t  (k  1) Tc
φ k (t)   Tc

 0 otherwise

 2
~  sin(2π f t) kT  t  (k  1) T
φ (t)   T c c c
k c

 0 otherwise

k  0,1,...... .......... .., N  1

where
Tc is chip duration,
N is number of chips per bit.

Transmitted signal x(t) for the interval of an information bit is

x(t)  c(t) s(t)

2Eb
 c(t) cos(2π fc t)
where, Eb isTbsignal energy per bit.
PN Code sequence
Eb N1 { c0, c1, ……cN-1} with ck= + 1
   k x(t)
Transmitted signal
N k 0
c φ 0  t  TN
k (t) is therefore b dimensional and requires N orthonormal

functions to represent it.

j(t) represent interfering signal (jammer). As said jammer tries to places all its
available energy in exactly same N dimension signal space. But jammer has no
knowledge of signal phase. Hence tries to place equal energy in two phase
coordinates that is cosine and sine
As per that jammer can be represented as
N 1 N1 ~~
j(t)   jkφk (t) 
k 0
j
k 0
k φk (t) 0  t  Tb

where
Tb

jk   j(t) φ
0
k (t) dt k  0,1,...... ........N  1

~ T ~b

j   j(t) φk (t) dt k  0,1,..............N  1

k 0

Thus j(t) is 2N dimensional, twice the dimension as that of x(t).

183
 Average interference power of j(t)
Tb
1
J j
2
(t) dt
Tb 0

N 1 2
~
N 1 2
1 1

Tb

k 0
j 
k Tb 
k 0
j
k

as jammer places equal energy in two phase coordinates , hence

N 1
~ N 1 2
2

 jk
k 0
 j
k 0
k

N 1
2 2
J
Tb
 jk
k 0

To evaluate system performance we calculate SNR at input and output of

DS/BPSK receiver.
The coherent receiver input is u(t) =s(t) + c(t)j(t)
and using this u(t), output at coherent receiver

Tb
2
v
Tb  u(t) cos(2π f t)dt
0
c

 v s  v cj

Where vs is despread component of BPSK and vcj of spread interference.

1
R c (k) 
N

Tb
2
v cj 
Tb  c(t) j(t) cos(2π f t)dt
0
c

Consider despread BPSK signal s(t)

2Eb
s(t)   cos(2π fc t) 0  t  Tb
Tb

184
 Where + sign is for symbol 1
- sign for symbol 0.
If carrier frequency is integer multiple of 1 / Tb , wehave
vs   Eb

Consider spread interference component vcj

here c(t) is considered in sequence form { c0, c1, ……cN-1}
Tb
Tc N1
v cj 
Tb
 ck  j(t)φk (t) dt
k 0 0

Tc N1

Tb
c
k 0
k
jk

With Ck treated as independent identical random variables with both symbols

having equal probabilities
1
P(Ck  1)  P(Ck   1) 
2

Expected value of Random variable vcj is zero, for fixed k we have

E Ck jk jk   jk P(Ck  1)  jkP(Ck   1)
1 1
 jk  jk
2 2
0

and Variance

 
Var Vc j j 
1 N 1 2
 jk
N k 0

JTc
2

Spread factor N = Tb/Tc

Output signal to noise ratio is
2Eb
(SNR)0 
JTc

The average signal power at receiver input is E b/Tb hence input SNR
Eb /Tb
(SNR)I 
J

185
 2Tb
(SNR)0  (SNR)I
Tc

Expressing SNR in decibels

10 log 10(SNR) 0  10 log 10(SNR)I  3  10 log 10(PG), dB

where Tb
PG 
Tc

 .3db term on right side accounts for gain in SNR due to coherent
detection.
 . Last term accounts for gain in SNR by use of spread spectrum.
PG is called Processing Gain

1. Bit rate of binary data entering the transmitter input is

1
Rb 
Tb

2. The bandwidth of PN sequence c(t) , of main lobe is W c

1
Wc 
Tc

Wc
PG 
Rb
186
Probability of error

To calculate probability of error, we consider output component v of coherent

detector as sample value of random variable V
V  Eb  Vcj

Eb is signal energy per bit and Vcj is noise component

Decision rule is, if detector output exceeds a threshold of zero volts; received bit
is symbol 1 else decision is favored for zero.

• Average probability of error Pe is nothing but conditional probability which

depends on random variable Vcj.

• As a result receiver makes decision in favor of symbol 1 when symbol 0

transmitted and vice versa
• Random variable Vcj is sum of N such random variables. Hence for
Large N it can assume Gaussian distribution .
• As mean and variance has already been discussed , zero mean and
variance JTc/2
Probability of error can be calculated from simple formula for DS/BPSK system

1  Eb 
Pe  erfc  

2  JTc 

Antijam Characteristics
Consider error probability of BPSK

1  Eb 
Pe  erfc  

2  N0 

Comparing both probabilities;

N0 JTc

2 2

Since bit energy Eb =PTb , P= average signal power.

We can express bit energy to noise density ratio as

Eb  Tb   P 
   
N0  Tc   J 

187
 J PG
or 
P Eb / N0

The ratio J/P is termed jamming margin. Jamming Margin is expressed in

decibels as  Eb 
(jamming mar gin) dB  (P r ocessing gain) dB  10log 10  
 N 0  min

 Eb 
 
 N0 
Where is minimum bit energy to noise ratio needed to support a
prescribed average probability of error.

Example1
A pseudo random sequence is generated using a feed back shift register of
length m=4. The chip rate is 107 chips per second. Find the following
a) PN sequence length
b) Chip duration of PN sequence
c) PN sequence period
Solution
a) Length of PN sequence N = 2m-1
= 24-1 =15
b) Chip duration Tc = 1/chip rate =1/107 = 0.1µsec
c) PN sequence period T = NTc
= 15 x 0.1µsec = 1.5µsec

Example2
A direct sequence spread binary phase shift keying system uses a
feedback shift register of length 19 for the generation of PN sequence.
Calculate the processing gain of the system.
Solution
Given length of shift register = m =19
Therefore length of PN sequence N = 2m - 1
= 219 - 1
Processing gain PG = Tb/Tc =N
in db =10log10N = 10 log10 (219)
= 57db
Example3
A Spread spectrum communication system has the following parameters.
Information bit duration Tb = 1.024 msecs and PN chip duration of 1µsecs.

188
The average probability of error of system is not to exceed 10 -5. calculate
a) Length of shift register b) Processing gain c) jamming margin
Solution
Processing gain PG =N= Tb/Tc =1024 corresponding length of shift register m =
10
In case of coherent BPSK For Probability of error 10 -5.
[Referring to error function table]
Eb/N0 =10.8
Therefore jamming margin
 Eb 
(jamming mar gin)dB  (Pr ocessing gain) dB  10log10  
 N0 min
 Eb 
(jamming margin)dB  10log10 PG dB  10log10  
 N 0  min

 10log101024  10log1010.8

= 30.10 – 10.33
= 19.8 db
Frequency – Hop Spread Spectrum:
In a frequency – hop Spread – Spectrum technique, the spectrum of data
modulated carrier is widened by changing the carrier frequency in a pseudo –
random manner. The type of spread – spectrum in which the carrier hops
randomly form one frequency to another is called Frequency – Hop (FH)
Spread Spectrum.
Since frequency hopping does not covers the entire spread spectrum
instantaneously. We are led to consider the rate at which the hop occurs.
Depending upon this we have two types of frequency hop.
1. Slow frequency hopping:- In which the symbol rate Rs of the MFSK signal
is an integer multiple of the hop rate Rh. That is several symbols are
transmitted on each frequency hop.
2. Fast – Frequency hopping:- In which the hop rate Rh is an integral multiple
of the MFSK symbol rate Rs. That is the carrier frequency will hoop
several times during the transmission of one symbol.
A common modulation format for frequency hopping system is that of
M- ary frequency – shift – keying (MFSK).

189
Slow frequency hopping:-

Fig.a) Shows the block diagram of an FH / MFSK transmitter, which involves

frequency modulation followed by mixing.
The incoming binary data are applied to an M-ary FSK modulator. The resulting
modulated wave and the output from a digital frequency synthesizer are then
applied to a mixer that consists of a multiplier followed by a band – pass filter.
The filter is designed to select the sum frequency component resulting from the
multiplication process as the transmitted signal. An ‘k’ bit segments of a PN
sequence drive the frequency synthesizer, which enables the carrier frequency to
hop over 2n distinct values. Since frequency synthesizers are unable to maintain
phase coherence over successive hops, most frequency hops spread spectrum
communication system use non coherent M-ary modulation system.

Fig a :- Frequency hop spread M-ary Frequency – shift – keying

In the receiver the frequency hoping is first removed by mixing the received
signal with the output of a local frequency synthesizer that is synchronized with
the transmitter. The resulting output is then band pass filtered and subsequently
processed by a non coherent M-ary FSK demodulator. To implement this M-ary

190
detector, a bank of M non coherent matched filters, each of which is matched to
one of the MFSK tones is used. By selecting the largest filtered output, the
original transmitted signal is estimated.
An individual FH / MFSK tone of shortest duration is referred as a chip. The chip
rate Rc for an FH / MFSK system is defined by
Rc = Max(Rh,Rs)
Where Rh is the hop rate and Rs is Symbol Rate
In a slow rate frequency hopping multiple symbols are transmitted per hop.
Hence each symbol of a slow FH / MFSK signal is a chip. The bit rate R b of the
incoming binary data. The symbol rate Rs of the MFSK signal, the chip rate Rc
and the hop rate Rn are related by
Rc = Rs = Rb /k ≥ Rh
where k= log2M
Fast frequency hopping:-
A fast FH / MFSK system differs from a slow FH / MFSK system in that
there are multiple hops per m-ary symbol. Hence in a fast FH / MFSK system
each hop is a chip.

Fast Frequency Hopping Slow Frequency Hopping

Several frequency hops Several modulation symbols per

Per modulation hop

Shortest uninterrupted waveform Shortest uninterrupted waveform

in the system is that of hop in the system is that of data
symbol

Chip duration =hop duration Chip duration=bit duration.

191
Fig. illustrates the variation of the frequency of a slow FH/MFSK signal with time
for one complete period of the PN sequence. The period of the PN sequence is
24-1 = 15. The FH/MFSK signal has the following parameters:
Number of bits per MFSK symbol K = 2.
Number of MFSK tones M = 2K = 4
Length of PN segment per hop k=3
Total number of frequency hops 2k = 8

192
Fig. illustrates the variation of the transmitted frequency of a fast FH/MFSK signal
with time. The signal has the following parameters:
Number of bits per MFSK symbol K = 2.
Number of MFSK tones M = 2K = 4
Length of PN segment per hop k=3
Total number of frequency hops 2k = 8

193
194